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quarto/integrals/partial_fractions.qmd

@@ -109,7 +109,7 @@ What remains is to establish that we can take $A(x) = a(x)\cdot P(x)$ with a deg
 In Proposition 3.8 of [Bradley](http://www.m-hikari.com/imf/imf-2012/29-32-2012/cookIMF29-32-2012.pdf) and Cook we can see how. Recall the division algorithm, for example, says there are $q_k$ and $r_k$ with $A=q\cdot q_k + r_k$ where the degree of $r_k$ is less than that of $q$, which is linear or quadratic. This is repeatedly applied below:
 
 
-$$
+
 \begin{align*}
 \frac{A}{q^k} &= \frac{q\cdot q_k + r_k}{q^k}\\
 &= \frac{r_k}{q^k} + \frac{q_k}{q^{k-1}}\\
@@ -119,7 +119,7 @@ $$
 &= \cdots\\
 &= \frac{r_k}{q^k} + \frac{r_{k-1}}{q^{k-1}} + \cdots + q_1.
 \end{align*}
-$$
+
 
 So the term $A(x)/q(x)^k$ can be expressed in terms of a sum where the numerators or each term have degree less than $q(x)$, as expected by the statement of the theorem.
 
@@ -208,14 +208,14 @@ integrate(B/((a*x)^2 - 1)^4, x)
 In [Bronstein](http://www-sop.inria.fr/cafe/Manuel.Bronstein/publications/issac98.pdf) 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 ($j>1$ and $b^2 - 4c < 0$ below):
 
 
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+
 \begin{align*}
 \int \frac{A}{(x-a)^j} &= \frac{A}{1-j}\frac{1}{(x-a)^{1-j}}\\
 \int \frac{A}{x-a}     &= A\log(x-a)\\
 \int \frac{Bx+C}{x^2 + bx + c}     &= \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} &= \frac{B' x + C'}{(x^2 + bx + c)^{j-1}} + \int \frac{C''}{(x^2 + bx + c)^{j-1}}
 \end{align*}
-$$
+
 
 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;
 
@@ -482,12 +482,12 @@ How to see that these give rise to real answers on integration is the point of t
 Breaking the terms up over $a$ and $b$ we have:
 
 
-$$
+
 \begin{align*}
 I &= \frac{a}{x - (\alpha + i \beta)} + \frac{a}{x - (\alpha - i \beta)} \\
 II &= i\frac{b}{x - (\alpha + i \beta)} - i\frac{b}{x - (\alpha - i \beta)}
 \end{align*}
-$$
+
 
 Integrating $I$ leads to two logarithmic terms, which are combined to give: