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<ul>
<li><a href="#partial-fraction-decomposition" id="toc-partial-fraction-decomposition" class="nav-link active" data-scroll-target="#partial-fraction-decomposition"> <span class="header-section-number">40.1</span> Partial fraction decomposition</a>
<ul class="collapse">
<li><a href="#sketch-of-proof" id="toc-sketch-of-proof" class="nav-link" data-scroll-target="#sketch-of-proof"> <span class="header-section-number">40.1.1</span> Sketch of proof</a></li>
</ul></li>
<li><a href="#integrating-the-terms-in-a-partial-fraction-decomposition" id="toc-integrating-the-terms-in-a-partial-fraction-decomposition" class="nav-link" data-scroll-target="#integrating-the-terms-in-a-partial-fraction-decomposition"> <span class="header-section-number">40.2</span> Integrating the terms in a partial fraction decomposition</a>
<ul class="collapse">
<li><a href="#linear-factors" id="toc-linear-factors" class="nav-link" data-scroll-target="#linear-factors"> <span class="header-section-number">40.2.1</span> Linear factors</a></li>
<li><a href="#quadratic-factors" id="toc-quadratic-factors" class="nav-link" data-scroll-target="#quadratic-factors"> <span class="header-section-number">40.2.2</span> Quadratic factors</a></li>
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<li><a href="#questions" id="toc-questions" class="nav-link" data-scroll-target="#questions"> <span class="header-section-number">40.3</span> Questions</a></li>
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<h1 class="title d-none d-lg-block"><span class="chapter-number">40</span>&nbsp; <span class="chapter-title">Partial Fractions</span></h1>
</div>
<div class="quarto-title-meta">
</div>
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<div class="sourceCode cell-code" id="cb1"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb1-1"><a href="#cb1-1" aria-hidden="true" tabindex="-1"></a><span class="im">using</span> <span class="bu">CalculusWithJulia</span></span>
<span id="cb1-2"><a href="#cb1-2" aria-hidden="true" tabindex="-1"></a><span class="im">using</span> <span class="bu">SymPy</span></span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<hr>
<p>Integration is facilitated when an antiderivative for <span class="math inline">\(f\)</span> can be found, as then definite integrals can be evaluated through the fundamental theorem of calculus.</p>
<p>However, despite integration being an algorithmic procedure, integration is not. There are “tricks” to try, such as substitution and integration by parts. These work in some cases. However, there are classes of functions for which algorithms exist. For example, the <code>SymPy</code> <code>integrate</code> function mostly implements an algorithm that decides if an elementary function has an antiderivative. The <a href="http://en.wikipedia.org/wiki/Elementary_function">elementary</a> functions include exponentials, their inverses (logarithms), trigonometric functions, their inverses, and powers, including <span class="math inline">\(n\)</span>th roots. Not every elementary function will have an antiderivative comprised of (finite) combinations of elementary functions. The typical example is <span class="math inline">\(e^{x^2}\)</span>, which has no simple antiderivative, despite its ubiquitousness.</p>
<p>There are classes of functions where an (elementary) antiderivative can always be found. Polynomials provide a case. More surprisingly, so do their ratios, <em>rational functions</em>.</p>
<section id="partial-fraction-decomposition" class="level2" data-number="40.1">
<h2 data-number="40.1" class="anchored" data-anchor-id="partial-fraction-decomposition"><span class="header-section-number">40.1</span> Partial fraction decomposition</h2>
<p>Let <span class="math inline">\(f(x) = p(x)/q(x)\)</span>, where <span class="math inline">\(p\)</span> and <span class="math inline">\(q\)</span> are polynomial functions with real coefficients. Further, we assume without comment that <span class="math inline">\(p\)</span> and <span class="math inline">\(q\)</span> have no common factors. (If they did, we can divide them out, an act which has no effect on the integrability of <span class="math inline">\(f(x)\)</span>.</p>
<p>The function <span class="math inline">\(q(x)\)</span> will factor over the real numbers. The fundamental theorem of algebra can be applied to say that <span class="math inline">\(q(x)=q_1(x)^{n_1} \cdots q_k(x)^{n_k}\)</span> where <span class="math inline">\(q_i(x)\)</span> is a linear or quadratic polynomial and <span class="math inline">\(n_k\)</span> a positive integer.</p>
<blockquote class="blockquote">
<p><strong>Partial Fraction Decomposition</strong>: There are unique polynomials <span class="math inline">\(a_{ij}\)</span> with degree <span class="math inline">\(a_{ij} &lt;\)</span> degree <span class="math inline">\(q_i\)</span> such that</p>
<p><span class="math display">\[
\frac{p(x)}{q(x)} = a(x) + \sum_{i=1}^k \sum_{j=1}^{n_i} \frac{a_{ij}(x)}{q_i(x)^j}.
\]</span></p>
</blockquote>
<p>The method is attributed to John Bernoulli, one of the prolific Bernoulli brothers who put a stamp on several areas of math. This Bernoulli was a mentor to Euler.</p>
<p>This basically says that each factor <span class="math inline">\(q_i(x)^{n_i}\)</span> contributes a term like:</p>
<p><span class="math display">\[
\frac{a_{i1}(x)}{q_i(x)^1} + \frac{a_{i2}(x)}{q_i(x)^2} + \cdots + \frac{a_{in_i}(x)}{q_i(x)^{n_i}},
\]</span></p>
<p>where each <span class="math inline">\(a_{ij}(x)\)</span> has degree less than the degree of <span class="math inline">\(q_i(x)\)</span>.</p>
<p>The value of this decomposition is that the terms <span class="math inline">\(a_{ij}(x)/q_i(x)^j\)</span> each have an antiderivative, and so the sum of them will also have an antiderivative.</p>
<div class="callout-note callout callout-style-default callout-captioned">
<div class="callout-header d-flex align-content-center">
<div class="callout-icon-container">
<i class="callout-icon"></i>
</div>
<div class="callout-caption-container flex-fill">
Note
</div>
</div>
<div class="callout-body-container callout-body">
<p>Many calculus texts will give some examples for finding a partial fraction decomposition. We push that work off to <code>SymPy</code>, as for all but the easiest cases - a few are in the problems - it can be a bit tedious.</p>
</div>
</div>
<p>In <code>SymPy</code>, the <code>apart</code> function will find the partial fraction decomposition when a factorization is available. For example, here we see <span class="math inline">\(n_i\)</span> terms for each power of <span class="math inline">\(q_i\)</span></p>
<div class="cell" data-execution_count="4">
<div class="sourceCode cell-code" id="cb2"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb2-1"><a href="#cb2-1" aria-hidden="true" tabindex="-1"></a><span class="pp">@syms</span> a<span class="op">::</span><span class="dt">real </span>b<span class="op">::</span><span class="dt">real </span>c<span class="op">::</span><span class="dt">real </span>A<span class="op">::</span><span class="dt">real </span>B<span class="op">::</span><span class="dt">real </span>x<span class="op">::</span><span class="dt">real</span></span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="5">
<pre><code>(a, b, c, A, B, x)</code></pre>
</div>
</div>
<div class="cell" data-execution_count="5">
<div class="sourceCode cell-code" id="cb4"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb4-1"><a href="#cb4-1" aria-hidden="true" tabindex="-1"></a><span class="fu">apart</span>((x<span class="op">-</span><span class="fl">2</span>)<span class="fu">*</span>(x<span class="op">-</span><span class="fl">3</span>) <span class="op">/</span> (<span class="fu">x*</span>(x<span class="op">-</span><span class="fl">1</span>)<span class="op">^</span><span class="fl">2</span><span class="fu">*</span>(x<span class="op">^</span><span class="fl">2</span> <span class="op">+</span> <span class="fl">2</span>)<span class="op">^</span><span class="fl">3</span>))</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="6">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
- \frac{8 x - 13}{9 \left(x^{2} + 2\right)^{3}} - \frac{35 x - 34}{54 \left(x^{2} + 2\right)^{2}} - \frac{45 x - 28}{108 \left(x^{2} + 2\right)} - \frac{1}{3 \left(x - 1\right)} + \frac{2}{27 \left(x - 1\right)^{2}} + \frac{3}{4 x}
\]
</span>
</div>
</div>
<section id="sketch-of-proof" class="level3" data-number="40.1.1">
<h3 data-number="40.1.1" class="anchored" data-anchor-id="sketch-of-proof"><span class="header-section-number">40.1.1</span> Sketch of proof</h3>
<p>A standard proof uses two facts of number systems: the division algorithm and a representation of the greatest common divisor in terms of sums, extended to polynomials. Our sketch shows how these are used.</p>
<p>Take one of the factors of the denominators, and consider this representation of the rational function <span class="math inline">\(P(x)/(q(x)^k Q(x))\)</span> where there are no common factors to any of the three polynomials.</p>
<p>Since <span class="math inline">\(q(x)\)</span> and <span class="math inline">\(Q(x)\)</span> share no factors, <a href="http://tinyurl.com/kd6prns">Bezout’s</a> identity says there exists polynomials <span class="math inline">\(a(x)\)</span> and <span class="math inline">\(b(x)\)</span> with:</p>
<p><span class="math display">\[
a(x) Q(x) + b(x) q(x) = 1.
\]</span></p>
<p>Then dividing by <span class="math inline">\(q^k(x)Q(x)\)</span> gives the decomposition</p>
<p><span class="math display">\[
\frac{1}{q(x)^k Q(x)} = \frac{a(x)}{q(x)^k} + \frac{b(x)}{q(x)^{k-1}Q(x)}.
\]</span></p>
<p>So we get by multiplying the <span class="math inline">\(P(x)\)</span>:</p>
<p><span class="math display">\[
\frac{P(x)}{q(x)^k Q(x)} = \frac{A(x)}{q(x)^k} + \frac{B(x)}{q(x)^{k-1}Q(x)}.
\]</span></p>
<p>This may look more complicated, but what it does is peel off one term (The first) and leave something which is smaller, in this case by a factor of <span class="math inline">\(q(x)\)</span>. This process can be repeated pulling off a power of a factor at a time until nothing is left to do.</p>
<p>What remains is to establish that we can take <span class="math inline">\(A(x) = a(x)\cdot P(x)\)</span> with a degree less than that of <span class="math inline">\(q(x)\)</span>.</p>
<p>In Proposition 3.8 of <a href="http://www.m-hikari.com/imf/imf-2012/29-32-2012/cookIMF29-32-2012.pdf">Bradley</a> and Cook we can see how. Recall the division algorithm, for example, says there are <span class="math inline">\(q_k\)</span> and <span class="math inline">\(r_k\)</span> with <span class="math inline">\(A=q\cdot q_k + r_k\)</span> where the degree of <span class="math inline">\(r_k\)</span> is less than that of <span class="math inline">\(q\)</span>, which is linear or quadratic. This is repeatedly applied below:</p>
<p><span class="math display">\[
\begin{align*}
\frac{A}{q^k} &amp;= \frac{q\cdot q_k + r_k}{q^k}\\
&amp;= \frac{r_k}{q^k} + \frac{q_k}{q^{k-1}}\\
&amp;= \frac{r_k}{q^k} + \frac{q \cdot q_{k-1} + r_{k-1}}{q^{k-1}}\\
&amp;= \frac{r_k}{q^k} + \frac{r_{k-1}}{q^{k-1}} + \frac{q_{k-1}}{q^{k-2}}\\
&amp;= \frac{r_k}{q^k} + \frac{r_{k-1}}{q^{k-1}} + \frac{q\cdot q_{k-2} + r_{k-2}}{q^{k-2}}\\
&amp;= \cdots\\
&amp;= \frac{r_k}{q^k} + \frac{r_{k-1}}{q^{k-1}} + \cdots + q_1.
\end{align*}
\]</span></p>
<p>So the term <span class="math inline">\(A(x)/q(x)^k\)</span> can be expressed in terms of a sum where the numerators or each term have degree less than <span class="math inline">\(q(x)\)</span>, as expected by the statement of the theorem.</p>
</section>
</section>
<section id="integrating-the-terms-in-a-partial-fraction-decomposition" class="level2" data-number="40.2">
<h2 data-number="40.2" class="anchored" data-anchor-id="integrating-the-terms-in-a-partial-fraction-decomposition"><span class="header-section-number">40.2</span> Integrating the terms in a partial fraction decomposition</h2>
<p>We discuss, by example, how each type of possible term in a partial fraction decomposition has an antiderivative. Hence, rational functions will <em>always</em> have an antiderivative that can be computed.</p>
<section id="linear-factors" class="level3" data-number="40.2.1">
<h3 data-number="40.2.1" class="anchored" data-anchor-id="linear-factors"><span class="header-section-number">40.2.1</span> Linear factors</h3>
<p>For <span class="math inline">\(j=1\)</span>, if <span class="math inline">\(q_i\)</span> is linear, then <span class="math inline">\(a_{ij}/q_i^j\)</span> must look like a constant over a linear term, or something like:</p>
<div class="cell" data-execution_count="6">
<div class="sourceCode cell-code" id="cb5"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb5-1"><a href="#cb5-1" aria-hidden="true" tabindex="-1"></a>p <span class="op">=</span> a<span class="op">/</span>(x<span class="op">-</span>c)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="7">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
\frac{a}{- c + x}
\]
</span>
</div>
</div>
<p>This has a logarithmic antiderivative:</p>
<div class="cell" data-execution_count="7">
<div class="sourceCode cell-code" id="cb6"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb6-1"><a href="#cb6-1" aria-hidden="true" tabindex="-1"></a><span class="fu">integrate</span>(p, x)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="8">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
a \log{\left(- c + x \right)}
\]
</span>
</div>
</div>
<p>For <span class="math inline">\(j &gt; 1\)</span>, we have powers.</p>
<div class="cell" data-execution_count="8">
<div class="sourceCode cell-code" id="cb7"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb7-1"><a href="#cb7-1" aria-hidden="true" tabindex="-1"></a><span class="pp">@syms</span> j<span class="op">::</span><span class="dt">positive</span></span>
<span id="cb7-2"><a href="#cb7-2" aria-hidden="true" tabindex="-1"></a><span class="fu">integrate</span>(a<span class="op">/</span>(x<span class="op">-</span>c)<span class="op">^</span>j, x)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="9">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
a \left(\begin{cases} \frac{\left(- c + x\right)^{1 - j}}{1 - j} &amp; \text{for}\: j \neq 1 \\\log{\left(- c + x \right)} &amp; \text{otherwise} \end{cases}\right)
\]
</span>
</div>
</div>
</section>
<section id="quadratic-factors" class="level3" data-number="40.2.2">
<h3 data-number="40.2.2" class="anchored" data-anchor-id="quadratic-factors"><span class="header-section-number">40.2.2</span> Quadratic factors</h3>
<p>When <span class="math inline">\(q_i\)</span> is quadratic, it looks like <span class="math inline">\(ax^2 + bx + c\)</span>. Then <span class="math inline">\(a_{ij}\)</span> can be a constant or a linear polynomial. The latter can be written as <span class="math inline">\(Ax + B\)</span>.</p>
<p>The integral of the following general form is presented below:</p>
<p><span class="math display">\[
\frac{Ax +B }{(ax^2 + bx + c)^j},
\]</span></p>
<p>With <code>SymPy</code>, we consider a few cases of the following form, which results from a shift of <code>x</code></p>
<p><span class="math display">\[
\frac{Ax + B}{((ax)^2 \pm 1)^j}
\]</span></p>
<p>This can be done by finding a <span class="math inline">\(d\)</span> so that <span class="math inline">\(a(x-d)^2 + b(x-d) + c = dx^2 + e = e((\sqrt{d/e}x^2 \pm 1)\)</span>.</p>
<p>The integrals of the type <span class="math inline">\(Ax/((ax)^2 \pm 1)\)</span> can completed by <span class="math inline">\(u\)</span>-substitution, with <span class="math inline">\(u=(ax)^2 \pm 1\)</span>.</p>
<p>For example,</p>
<div class="cell" data-execution_count="9">
<div class="sourceCode cell-code" id="cb8"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb8-1"><a href="#cb8-1" aria-hidden="true" tabindex="-1"></a><span class="fu">integrate</span>(A<span class="op">*</span>x<span class="op">/</span>((a<span class="op">*</span>x)<span class="op">^</span><span class="fl">2</span> <span class="op">+</span> <span class="fl">1</span>)<span class="op">^</span><span class="fl">4</span>, x)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="10">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
- \frac{A}{6 a^{8} x^{6} + 18 a^{6} x^{4} + 18 a^{4} x^{2} + 6 a^{2}}
\]
</span>
</div>
</div>
<p>The integrals of the type <span class="math inline">\(B/((ax)^2\pm 1)\)</span> are completed by trigonometric substitution and various reduction formulas. They can get involved, but are tractable. For example:</p>
<div class="cell" data-execution_count="10">
<div class="sourceCode cell-code" id="cb9"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb9-1"><a href="#cb9-1" aria-hidden="true" tabindex="-1"></a><span class="fu">integrate</span>(B<span class="op">/</span>((a<span class="op">*</span>x)<span class="op">^</span><span class="fl">2</span> <span class="op">+</span> <span class="fl">1</span>)<span class="op">^</span><span class="fl">4</span>, x)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="11">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
B \left(\frac{15 a^{4} x^{5} + 40 a^{2} x^{3} + 33 x}{48 a^{6} x^{6} + 144 a^{4} x^{4} + 144 a^{2} x^{2} + 48} + \frac{5 \operatorname{atan}{\left(a x \right)}}{16 a}\right)
\]
</span>
</div>
</div>
<p>and</p>
<div class="cell" data-execution_count="11">
<div class="sourceCode cell-code" id="cb10"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb10-1"><a href="#cb10-1" aria-hidden="true" tabindex="-1"></a><span class="fu">integrate</span>(B<span class="op">/</span>((a<span class="op">*</span>x)<span class="op">^</span><span class="fl">2</span> <span class="op">-</span> <span class="fl">1</span>)<span class="op">^</span><span class="fl">4</span>, x)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="12">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
B \left(\frac{- 15 a^{4} x^{5} + 40 a^{2} x^{3} - 33 x}{48 a^{6} x^{6} - 144 a^{4} x^{4} + 144 a^{2} x^{2} - 48} - \frac{5 \log{\left(x - \frac{1}{a} \right)}}{32 a} + \frac{5 \log{\left(x + \frac{1}{a} \right)}}{32 a}\right)
\]
</span>
</div>
</div>
<hr>
<p>In <a href="http://www-sop.inria.fr/cafe/Manuel.Bronstein/publications/issac98.pdf">Bronstein</a> this characterization can be found - “This method, which dates back to Newton, Leibniz and Bernoulli, should not be used in practice, yet it remains the method found in most calculus texts and is often taught. Its major drawback is the factorization of the denominator of the integrand over the real or complex numbers.” We can also find the following formulas which formalize the above exploratory calculations (<span class="math inline">\(j&gt;1\)</span> and <span class="math inline">\(b^2 - 4c &lt; 0\)</span> below):</p>
<p><span class="math display">\[
\begin{align*}
\int \frac{A}{(x-a)^j} &amp;= \frac{A}{1-j}\frac{1}{(x-a)^{1-j}}\\
\int \frac{A}{x-a} &amp;= A\log(x-a)\\
\int \frac{Bx+C}{x^2 + bx + c} &amp;= \frac{B}{2} \log(x^2 + bx + c) + \frac{2C-bB}{\sqrt{4c-b^2}}\cdot \arctan\left(\frac{2x+b}{\sqrt{4c-b^2}}\right)\\
\int \frac{Bx+C}{(x^2 + bx + c)^j} &amp;= \frac{B' x + C'}{(x^2 + bx + c)^{j-1}} + \int \frac{C''}{(x^2 + bx + c)^{j-1}}
\end{align*}
\]</span></p>
<p>The first returns a rational function; the second yields a logarithm term; the third yields a logarithm and an arctangent term; while the last, which has explicit constants available, provides a reduction that can be recursively applied;</p>
<p>That is integrating <span class="math inline">\(f(x)/g(x)\)</span>, a rational function, will yield an output that looks like the following, where the functions are polynomials:</p>
<p><span class="math display">\[
\int f(x)/g(x) = P(x) + \frac{C(x)}{D{x}} + \sum v_i \log(V_i(x)) + \sum w_j \arctan(W_j(x))
\]</span></p>
<p>(Bronstein also sketches the modern method which is to use a Hermite reduction to express <span class="math inline">\(\int (f/g) dx = p/q + \int (g/h) dx\)</span>, where <span class="math inline">\(h\)</span> is square free (the “<code>j</code>” are all <span class="math inline">\(1\)</span>). The latter can be written over the complex numbers as logarithmic terms of the form <span class="math inline">\(\log(x-a)\)</span>, the “<code>a</code>s”found following a method due to Trager and Lazard, and Rioboo, which is mentioned in the SymPy documentation as the method used.)</p>
<section id="examples" class="level4">
<h4 class="anchored" data-anchor-id="examples">Examples</h4>
<p>Find an antiderivative for <span class="math inline">\(1/(x\cdot(x^2+1)^2)\)</span>.</p>
<p>We have a partial fraction decomposition is:</p>
<div class="cell" data-execution_count="12">
<div class="sourceCode cell-code" id="cb11"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb11-1"><a href="#cb11-1" aria-hidden="true" tabindex="-1"></a>q <span class="op">=</span> (x <span class="op">*</span> (x<span class="op">^</span><span class="fl">2</span> <span class="op">+</span> <span class="fl">1</span>)<span class="op">^</span><span class="fl">2</span>)</span>
<span id="cb11-2"><a href="#cb11-2" aria-hidden="true" tabindex="-1"></a><span class="fu">apart</span>(<span class="fl">1</span><span class="op">/</span>q)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="13">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
- \frac{x}{x^{2} + 1} - \frac{x}{\left(x^{2} + 1\right)^{2}} + \frac{1}{x}
\]
</span>
</div>
</div>
<p>We see three terms. The first and second will be done by <span class="math inline">\(u\)</span>-substitution, the third by a logarithm:</p>
<div class="cell" data-execution_count="13">
<div class="sourceCode cell-code" id="cb12"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb12-1"><a href="#cb12-1" aria-hidden="true" tabindex="-1"></a><span class="fu">integrate</span>(<span class="fl">1</span><span class="op">/</span>q, x)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="14">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
\log{\left(x \right)} - \frac{\log{\left(x^{2} + 1 \right)}}{2} + \frac{1}{2 x^{2} + 2}
\]
</span>
</div>
</div>
<hr>
<p>Find an antiderivative of <span class="math inline">\(1/(x^2 - 2x-3)\)</span>.</p>
<p>We again just let <code>SymPy</code> do the work. A partial fraction decomposition is given by:</p>
<div class="cell" data-execution_count="14">
<div class="sourceCode cell-code" id="cb13"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb13-1"><a href="#cb13-1" aria-hidden="true" tabindex="-1"></a>𝒒 <span class="op">=</span> (x<span class="op">^</span><span class="fl">2</span> <span class="op">-</span> <span class="fl">2</span>x <span class="op">-</span> <span class="fl">3</span>)</span>
<span id="cb13-2"><a href="#cb13-2" aria-hidden="true" tabindex="-1"></a><span class="fu">apart</span>(<span class="fl">1</span><span class="op">/</span>𝒒)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="15">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
- \frac{1}{4 \left(x + 1\right)} + \frac{1}{4 \left(x - 3\right)}
\]
</span>
</div>
</div>
<p>We see what should yield two logarithmic terms:</p>
<div class="cell" data-execution_count="15">
<div class="sourceCode cell-code" id="cb14"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb14-1"><a href="#cb14-1" aria-hidden="true" tabindex="-1"></a><span class="fu">integrate</span>(<span class="fl">1</span><span class="op">/</span>𝒒, x)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="16">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
\frac{\log{\left(x - 3 \right)}}{4} - \frac{\log{\left(x + 1 \right)}}{4}
\]
</span>
</div>
</div>
<div class="callout-note callout callout-style-default callout-captioned">
<div class="callout-header d-flex align-content-center">
<div class="callout-icon-container">
<i class="callout-icon"></i>
</div>
<div class="callout-caption-container flex-fill">
Note
</div>
</div>
<div class="callout-body-container callout-body">
<p><code>SymPy</code> will find <span class="math inline">\(\log(x)\)</span> as an antiderivative for <span class="math inline">\(1/x\)</span>, but more generally, <span class="math inline">\(\log(\lvert x\rvert)\)</span> is one.</p>
</div>
</div>
<section id="example" class="level5">
<h5 class="anchored" data-anchor-id="example">Example</h5>
<p>The answers found can become quite involved. <a href="https://arxiv.org/pdf/1712.01752.pdf">Corless</a>, Moir, Maza, and Xie use this example which at first glance seems tame enough:</p>
<div class="cell" data-execution_count="16">
<div class="sourceCode cell-code" id="cb15"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb15-1"><a href="#cb15-1" aria-hidden="true" tabindex="-1"></a>ex <span class="op">=</span> (x<span class="op">^</span><span class="fl">2</span> <span class="op">-</span> <span class="fl">1</span>) <span class="op">/</span> (x<span class="op">^</span><span class="fl">4</span> <span class="op">+</span> <span class="fl">5</span>x<span class="op">^</span><span class="fl">2</span> <span class="op">+</span> <span class="fl">7</span>)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="17">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
\frac{x^{2} - 1}{x^{4} + 5 x^{2} + 7}
\]
</span>
</div>
</div>
<p>But the integral is something best suited to a computer algebra system:</p>
<div class="cell" data-execution_count="17">
<div class="sourceCode cell-code" id="cb16"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb16-1"><a href="#cb16-1" aria-hidden="true" tabindex="-1"></a><span class="fu">integrate</span>(ex, x)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="18">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
\sqrt{\frac{17}{84} + \frac{13 \sqrt{7}}{168}} \log{\left(x^{2} + x \left(- \frac{38 \sqrt{6} \sqrt{34 + 13 \sqrt{7}}}{9} - \frac{1301 \sqrt{42} \sqrt{34 + 13 \sqrt{7}}}{1638} + \frac{19 \sqrt{42} \sqrt{34 + 13 \sqrt{7}} \sqrt{884 \sqrt{7} + 2339}}{546}\right) - \frac{2124092 \sqrt{884 \sqrt{7} + 2339}}{31941} - \frac{9481 \sqrt{7} \sqrt{884 \sqrt{7} + 2339}}{378} + \frac{290246555}{63882} + \frac{4221850 \sqrt{7}}{2457} \right)} - \sqrt{\frac{17}{84} + \frac{13 \sqrt{7}}{168}} \log{\left(x^{2} + x \left(- \frac{19 \sqrt{42} \sqrt{34 + 13 \sqrt{7}} \sqrt{884 \sqrt{7} + 2339}}{546} + \frac{1301 \sqrt{42} \sqrt{34 + 13 \sqrt{7}}}{1638} + \frac{38 \sqrt{6} \sqrt{34 + 13 \sqrt{7}}}{9}\right) - \frac{2124092 \sqrt{884 \sqrt{7} + 2339}}{31941} - \frac{9481 \sqrt{7} \sqrt{884 \sqrt{7} + 2339}}{378} + \frac{290246555}{63882} + \frac{4221850 \sqrt{7}}{2457} \right)} + 2 \sqrt{- \frac{\sqrt{884 \sqrt{7} + 2339}}{84} + \frac{17}{84} + \frac{13 \sqrt{7}}{56}} \operatorname{atan}{\left(\frac{78 \sqrt{42} x}{- 9 \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}} + 19 \sqrt{884 \sqrt{7} + 2339} \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}}} - \frac{988 \sqrt{7} \sqrt{34 + 13 \sqrt{7}}}{- 9 \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}} + 19 \sqrt{884 \sqrt{7} + 2339} \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}}} - \frac{1301 \sqrt{34 + 13 \sqrt{7}}}{- 9 \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}} + 19 \sqrt{884 \sqrt{7} + 2339} \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}}} + \frac{57 \sqrt{34 + 13 \sqrt{7}} \sqrt{884 \sqrt{7} + 2339}}{- 9 \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}} + 19 \sqrt{884 \sqrt{7} + 2339} \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}}} \right)} + 2 \sqrt{- \frac{\sqrt{884 \sqrt{7} + 2339}}{84} + \frac{17}{84} + \frac{13 \sqrt{7}}{56}} \operatorname{atan}{\left(\frac{78 \sqrt{42} x}{- 9 \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}} + 19 \sqrt{884 \sqrt{7} + 2339} \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}}} - \frac{57 \sqrt{34 + 13 \sqrt{7}} \sqrt{884 \sqrt{7} + 2339}}{- 9 \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}} + 19 \sqrt{884 \sqrt{7} + 2339} \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}}} + \frac{1301 \sqrt{34 + 13 \sqrt{7}}}{- 9 \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}} + 19 \sqrt{884 \sqrt{7} + 2339} \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}}} + \frac{988 \sqrt{7} \sqrt{34 + 13 \sqrt{7}}}{- 9 \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}} + 19 \sqrt{884 \sqrt{7} + 2339} \sqrt{- 2 \sqrt{884 \sqrt{7} + 2339} + 34 + 39 \sqrt{7}}} \right)}
\]
</span>
</div>
</div>
</section>
</section>
</section>
</section>
<section id="questions" class="level2" data-number="40.3">
<h2 data-number="40.3" class="anchored" data-anchor-id="questions"><span class="header-section-number">40.3</span> Questions</h2>
<section id="question" class="level6">
<h6 class="anchored" data-anchor-id="question">Question</h6>
<p>The partial fraction decomposition of <span class="math inline">\(1/(x(x-1))\)</span> must be of the form <span class="math inline">\(A/x + B/(x-1)\)</span>.</p>
<p>What is <span class="math inline">\(A\)</span>? (Use <code>SymPy</code> or just put the sum over a common denominator and solve for <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span>.)</p>
<div class="cell" data-hold="true" data-execution_count="18">
<div class="cell-output cell-output-display" data-execution_count="19">
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<p>What is <span class="math inline">\(B\)</span>?</p>
<div class="cell" data-hold="true" data-execution_count="19">
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<section id="question-1" class="level6">
<h6 class="anchored" data-anchor-id="question-1">Question</h6>
<p>The following gives the partial fraction decomposition for a rational expression:</p>
<p><span class="math display">\[
\frac{3x+5}{(1-2x)^2} = \frac{A}{1-2x} + \frac{B}{(1-2x)^2}.
\]</span></p>
<p>Find <span class="math inline">\(A\)</span> (being careful with the sign):</p>
<div class="cell" data-hold="true" data-execution_count="20">
<div class="cell-output cell-output-display" data-execution_count="21">
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<p>Find <span class="math inline">\(B\)</span>:</p>
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</section>
<section id="question-2" class="level6">
<h6 class="anchored" data-anchor-id="question-2">Question</h6>
<p>The following specifies the general partial fraction decomposition for a rational expression:</p>
<p><span class="math display">\[
\frac{1}{(x+1)(x-1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2}.
\]</span></p>
<p>Find <span class="math inline">\(A\)</span>:</p>
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<p>Find <span class="math inline">\(B\)</span>:</p>
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<p>Find <span class="math inline">\(C\)</span>:</p>
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</section>
<section id="question-3" class="level6">
<h6 class="anchored" data-anchor-id="question-3">Question</h6>
<p>Compute the following exactly:</p>
<p><span class="math display">\[
\int_0^1 \frac{(x-2)(x-3)}{(x-4)^2\cdot(x-5)} dx
\]</span></p>
<p>Is <span class="math inline">\(-6\log(5) - 5\log(3) - 1/6 + 11\log(4)\)</span> the answer?</p>
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</section>
<section id="question-4" class="level6">
<h6 class="anchored" data-anchor-id="question-4">Question</h6>
<p>In the assumptions for the partial fraction decomposition is the fact that <span class="math inline">\(p(x)\)</span> and <span class="math inline">\(q(x)\)</span> share no common factors. Suppose, this isn’t the case and in fact we have:</p>
<p><span class="math display">\[
\frac{p(x)}{q(x)} = \frac{(x-c)^m s(x)}{(x-c)^n t(x)}.
\]</span></p>
<p>Here <span class="math inline">\(s\)</span> and <span class="math inline">\(t\)</span> are polynomials such that <span class="math inline">\(s(c)\)</span> and <span class="math inline">\(t(c)\)</span> are non-zero.</p>
<p>If <span class="math inline">\(m &gt; n\)</span>, then why can we cancel out the <span class="math inline">\((x-c)^n\)</span> and not have a concern?</p>
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<code>SymPy</code> allows it.
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The value \(c\) is a removable singularity, so the integral will be identical.
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The resulting function has an identical domain and is equivalent for all \(x\).
</span>
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<p>If <span class="math inline">\(m = n\)</span>, then why can we cancel out the <span class="math inline">\((x-c)^n\)</span> and not have a concern?</p>
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<code>SymPy</code> allows it.
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The value \(c\) is a removable singularity, so the integral will be identical.
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The resulting function has an identical domain and is equivalent for all \(x\).
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<p>If <span class="math inline">\(m &lt; n\)</span>, then why can we cancel out the <span class="math inline">\((x-c)^n\)</span> and not have a concern?</p>
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The value \(c\) is a removable singularity, so the integral will be identical.
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The resulting function has an identical domain and is equivalent for all \(x\).
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<section id="question-5" class="level5">
<h5 class="anchored" data-anchor-id="question-5">Question</h5>
<p>The partial fraction decomposition, as presented, factors the denominator polynomial into linear and quadratic factors over the real numbers. Alternatively, factoring over the complex numbers is possible, resulting in terms like:</p>
<p><span class="math display">\[
\frac{a + ib}{x - (\alpha + i \beta)} + \frac{a - ib}{x - (\alpha - i \beta)}
\]</span></p>
<p>How to see that these give rise to real answers on integration is the point of this question.</p>
<p>Breaking the terms up over <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span> we have:</p>
<p><span class="math display">\[
\begin{align*}
I &amp;= \frac{a}{x - (\alpha + i \beta)} + \frac{a}{x - (\alpha - i \beta)} \\
II &amp;= i\frac{b}{x - (\alpha + i \beta)} - i\frac{b}{x - (\alpha - i \beta)}
\end{align*}
\]</span></p>
<p>Integrating <span class="math inline">\(I\)</span> leads to two logarithmic terms, which are combined to give:</p>
<p><span class="math display">\[
\int I dx = a\cdot \log((x-(\alpha+i\beta)) \cdot (x - (\alpha-i\beta)))
\]</span></p>
<p>This involves no complex numbers, as:</p>
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The \(\beta\) are \(0\), as the polynomials in question are real
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The complex numbers are complex conjugates, so the term in the logarithm will simply be \(x - 2\alpha x + \alpha^2 + \beta^2\)
</span>
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<p>The term <span class="math inline">\(II\)</span> benefits from this computation (attributed to Rioboo by <a href="https://arxiv.org/pdf/1712.01752.pdf">Corless et. al</a>)</p>
<p><span class="math display">\[
\frac{d}{dx} i \log(\frac{X+iY}{X-iY}) = 2\frac{d}{dx}\arctan(\frac{X}{Y})
\]</span></p>
<p>Applying this with <span class="math inline">\(X=x - \alpha\)</span> and <span class="math inline">\(Y=-\beta\)</span> shows that <span class="math inline">\(\int II dx\)</span> will be</p>
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\(-2b\arctan((x - \alpha)/(\beta))\)
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\(2b\sec^2(-(x-\alpha)/(-\beta))\)
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