Step By Step Calculus » 15.7 - Integration by Partial Fractions

Synopsis
The utility of partial fractions is that every proper rational function \dfrac{n(x)}{d(x)}\dfrac{n(x)}{d(x)} is representable as
 \displaystyle \frac{n(x)}{d(x)} \displaystyle \frac{n(x)}{d(x)} \displaystyle =\displaystyle = \displaystyle \sum\limits_{i=1}^{k}\left[ \frac{C_{i}^{1}}{b_{i}x+c_{i}}+\frac{C_{i}^{2}}{(b_{i}x+c_{i})^{2}}+\cdots +\frac{C_{i}^{r_{i}}}{(b_{i}x+c_{i})^{r_{i}}}\right] \displaystyle \sum\limits_{i=1}^{k}\left[ \frac{C_{i}^{1}}{b_{i}x+c_{i}}+\frac{C_{i}^{2}}{(b_{i}x+c_{i})^{2}}+\cdots +\frac{C_{i}^{r_{i}}}{(b_{i}x+c_{i})^{r_{i}}}\right] \displaystyle \displaystyle \displaystyle +\sum\limits_{i=k+1}^{k+l}\left[ \frac{B_{i}^{1}x+C_{i}^{1}}{a_{i}x^{2}+b_{i}x+c_{i}}+\frac{B_{i}^{2}x+C_{i}^{2}}{(a_{i}x^{2}+b_{i}x+c_{i})^{2}}+\cdots +\frac{B_{i}^{r_{i}}x+C_{i}^{r_{i}}}{(a_{i}x^{2}+b_{i}x+c_{i})^{r_{i}}}\right] \displaystyle +\sum\limits_{i=k+1}^{k+l}\left[ \frac{B_{i}^{1}x+C_{i}^{1}}{a_{i}x^{2}+b_{i}x+c_{i}}+\frac{B_{i}^{2}x+C_{i}^{2}}{(a_{i}x^{2}+b_{i}x+c_{i})^{2}}+\cdots +\frac{B_{i}^{r_{i}}x+C_{i}^{r_{i}}}{(a_{i}x^{2}+b_{i}x+c_{i})^{r_{i}}}\right]
(the partial fraction representation), where
\displaystyle d(x)=\prod\limits_{i=1}^{k}(b_{i}x+c_{i})^{r_{i}}\prod\limits_{i=k+1}^{k+l}(a_{i}x^{2}+b_{i}x+c_{i})^{r_{i}}.
\displaystyle d(x)=\prod\limits_{i=1}^{k}(b_{i}x+c_{i})^{r_{i}}\prod\limits_{i=k+1}^{k+l}(a_{i}x^{2}+b_{i}x+c_{i})^{r_{i}}.
This expansion can be integrated term by term. To integrate each term of the form
\displaystyle \frac{C_{i}^{k}}{(b_{i}x+c_{i})^{k}},
\displaystyle \frac{C_{i}^{k}}{(b_{i}x+c_{i})^{k}},
we use Forms A and B from the previous section to find that
\displaystyle \int \frac{C_{i}^{k}}{(b_{i}x+c_{i})^{k}}dx=\left\{ \begin{array}{ll} \frac{C_{i}^{k1}}{b_{i}}\ln \left| b_{i}x+c_{i}\right| +c & \text{if }k=1 \\ \frac{C_{i}^{k}}{b_{i}(1-k)}(b_{i}x+c_{i})^{1-k} & \text{if }k=2,3,... \end{array} \right.
\displaystyle \int \frac{C_{i}^{k}}{(b_{i}x+c_{i})^{k}}dx=\left\{ \begin{array}{ll} \frac{C_{i}^{k1}}{b_{i}}\ln \left| b_{i}x+c_{i}\right| +c & \text{if }k=1 \\ \frac{C_{i}^{k}}{b_{i}(1-k)}(b_{i}x+c_{i})^{1-k} & \text{if }k=2,3,... \end{array} \right.
Moreover, to integrate each term of the form
\displaystyle \frac{B_{i}^{k}x+C_{i}^{k}}{(a_{i}x^{2}+b_{i}x+c_{i})^{k}}
\displaystyle \frac{B_{i}^{k}x+C_{i}^{k}}{(a_{i}x^{2}+b_{i}x+c_{i})^{k}}
(with a_{i}>0a_{i}>0), we complete the square on the denominator by making the substitution u=\sqrt{a}x+\dfrac{b}{2\sqrt{a}}u=\sqrt{a}x+\dfrac{b}{2\sqrt{a}} so du=\sqrt{a}dxdu=\sqrt{a}dx and ax^{2}+bx+c=u^{2}+c-\dfrac{b^{2}}{4a}ax^{2}+bx+c=u^{2}+c-\dfrac{b^{2}}{4a}. Noting that d_{i}^{2}\circeq c_{i}-\dfrac{b_{i}^{2}}{4a_{i}}>0d_{i}^{2}\circeq c_{i}-\dfrac{b_{i}^{2}}{4a_{i}}>0 or else a_{i}x^{2}+b_{i}x+c_{i}a_{i}x^{2}+b_{i}x+c_{i} would factor, we have that
 \displaystyle \int \frac{B_{i}^{r_{i}}x+C_{i}^{r_{i}}}{(a_{i}x^{2}+b_{i}x+c_{i})^{r_{i}}}dx \displaystyle \int \frac{B_{i}^{r_{i}}x+C_{i}^{r_{i}}}{(a_{i}x^{2}+b_{i}x+c_{i})^{r_{i}}}dx \displaystyle =\displaystyle = \displaystyle \frac{1}{\sqrt{a_{i}}}\int \frac{B_{i}^{r_{i}}\left( \frac{u}{\sqrt{a_{i}}}-\frac{b_{i}}{2a_{i}}\right) +C_{i}^{r_{i}}}{(u^{2}+c_{i}-\frac{b_{i}^{2}}{4a_{i}})^{r_{i}}}du \displaystyle \frac{1}{\sqrt{a_{i}}}\int \frac{B_{i}^{r_{i}}\left( \frac{u}{\sqrt{a_{i}}}-\frac{b_{i}}{2a_{i}}\right) +C_{i}^{r_{i}}}{(u^{2}+c_{i}-\frac{b_{i}^{2}}{4a_{i}})^{r_{i}}}du \displaystyle \displaystyle \displaystyle =\displaystyle = \displaystyle \newcommand{\stackunder}[2]{\mathrel{\mathop{#2}\limits_{#1}}}\frac{B_{i}^{r_{i}}}{a_{i}}\int \frac{u}{(u^{2}+\stackunder{d_{i}^{2}}{\underbrace{c_{i}-\frac{b_{i}^{2}}{4a_{i}}}})^{r_{i}}}du+\frac{1}{\sqrt{a_{i}}}\int \frac{C_{i}^{r_{i}}-B_{i}^{r_{i}}\frac{b_{i}}{2a_{i}}}{(u^{2}+\stackunder{d_{i}^{2}}{\underbrace{c_{i}-\frac{b_{i}^{2}}{4a_{i}}}})^{r_{i}}}du. \displaystyle \newcommand{\stackunder}[2]{\mathrel{\mathop{#2}\limits_{#1}}}\frac{B_{i}^{r_{i}}}{a_{i}}\int \frac{u}{(u^{2}+\stackunder{d_{i}^{2}}{\underbrace{c_{i}-\frac{b_{i}^{2}}{4a_{i}}}})^{r_{i}}}du+\frac{1}{\sqrt{a_{i}}}\int \frac{C_{i}^{r_{i}}-B_{i}^{r_{i}}\frac{b_{i}}{2a_{i}}}{(u^{2}+\stackunder{d_{i}^{2}}{\underbrace{c_{i}-\frac{b_{i}^{2}}{4a_{i}}}})^{r_{i}}}du.
However, we use Forms A and B of the previous section to find that
\displaystyle \frac{B_{i}^{r_{i}}}{a_{i}}\int \frac{u}{(u^{2}+d_{i}^{2})^{r_{i}}}du=\left\{ \begin{array}{ll} \frac{B_{i}^{r_{i}}}{2a_{i}}\ln \left| u^{2}+d_{i}^{2}\right| +c & \text{if }r_{i}=1 \\ \frac{B_{i}^{r_{i}}}{2a_{i}(1-r_{i})}(u^{2}+d_{i}^{2})^{1-r_{i}} & \text{if }r_{i}=2,3,\ldots \end{array} \right.
\displaystyle \frac{B_{i}^{r_{i}}}{a_{i}}\int \frac{u}{(u^{2}+d_{i}^{2})^{r_{i}}}du=\left\{ \begin{array}{ll} \frac{B_{i}^{r_{i}}}{2a_{i}}\ln \left| u^{2}+d_{i}^{2}\right| +c & \text{if }r_{i}=1 \\ \frac{B_{i}^{r_{i}}}{2a_{i}(1-r_{i})}(u^{2}+d_{i}^{2})^{1-r_{i}} & \text{if }r_{i}=2,3,\ldots \end{array} \right.
Moreover,
\displaystyle \frac{1}{\sqrt{a_{i}}}\int \frac{C_{i}^{r_{i}}-B_{i}^{r_{i}}\frac{b_{i}}{2a_{i}}}{(u^{2}+d_{i}^{2})^{r_{i}}}du=\frac{C_{i}^{r_{i}}-B_{i}^{r_{i}}\frac{b_{i}}{2a_{i}}}{\sqrt{a_{i}}}I_{r},
\displaystyle \frac{1}{\sqrt{a_{i}}}\int \frac{C_{i}^{r_{i}}-B_{i}^{r_{i}}\frac{b_{i}}{2a_{i}}}{(u^{2}+d_{i}^{2})^{r_{i}}}du=\frac{C_{i}^{r_{i}}-B_{i}^{r_{i}}\frac{b_{i}}{2a_{i}}}{\sqrt{a_{i}}}I_{r},
where I_{r}\circeq \displaystyle\int \dfrac{1}{(u^{2}+d^{2})^{r}}duI_{r}\circeq \displaystyle\int \dfrac{1}{(u^{2}+d^{2})^{r}}du for r=1,2,\ldotsr=1,2,\ldots satisfies the reduction formula
 \displaystyle I_{r}=\frac{1}{d^{2}(2r-2)}\left[ \frac{u}{(u^{2}+d^{2})^{r-1}}+(2r-3)I_{r-1}\right] \displaystyle I_{r}=\frac{1}{d^{2}(2r-2)}\left[ \frac{u}{(u^{2}+d^{2})^{r-1}}+(2r-3)I_{r-1}\right] for \displaystyle r=2,3,.... \displaystyle r=2,3,....
subject to
\displaystyle I_{1}=\frac{1}{d_{i}}\tan ^{-1}(\frac{u}{d_{i}})+c.
\displaystyle I_{1}=\frac{1}{d_{i}}\tan ^{-1}(\frac{u}{d_{i}})+c.
Some other integrals can be converted to the integral of a rational function using a change of variables.