Question

Integrate the rational function: $\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}$

Answer 1

$\int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx$
$\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}$
Let $t=x^2$
$=\frac{(t+1)(t+2)}{(t+3)(t+4)}$
$=\frac{t^2+3t+2}{t^2+7t+12}$
$=\frac{t^2+7t+12-(4t+10)}{t^2+7t+12}$
$=1-\frac{4t+10}{t^2+7t+12}$
Using partial fraction
$\frac{4t+10}{(t+3)(t+4)}=\frac{A}{t+3}+\frac{B}{t+4}$
$\Rightarrow 4t+10=A(t+4)+B(t+3)$
Putting $t=-4$
$\Rightarrow 4,-4+10=A,-4+4,+B,-4+3$
$\Rightarrow -16+10=B,-1$
$\Rightarrow B=6$
Putting $t=-3$
$\Rightarrow 4(-3)+10$
$=A(-3+4)+B(-3+3)$
$\Rightarrow -12+10=A\times 1$
$\Rightarrow A=-2$
Putting $A$ and $B$
$\frac{4t+10}{(t-3)(t+4)}=\frac{-2}{t+3}+\frac{6}{t+4}$
Now,
$\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=1-\frac{4t+10}{t^2+7t+12}$
$=1-[\frac{-2}{t+3}+\frac{6}{t+4}]$
$=1-[\frac{-2}{x^2+3}+\frac{6}{x^2+4}]$
$=1+\frac{2}{x^2+3}-\frac{6}{x^2+4}$
Therefore,
$\int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}.dx$
$=\int (1+\frac{2}{x^2+3}-\frac{6}{x^2+4}).dx$
$=\int 1.dx+2\int \frac{1}{x^2+3}dx-6\int \frac{1}{x^2+4}dx$
$=\int 1.dx+2\int \frac{1}{x^2+(\sqrt{3}{)}^2}dx-6\int \frac{1}{x^2+2^2}dx$
$=x+2\left[\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{x}{\sqrt{3}}\right]-\left[\frac{1}{2}{\tan }^{-1}\frac{x}{2}\right]+C$
$[\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\frac{x}{a}]$
$=x+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)-3{\tan }^{-1}\left(\frac{x}{2}\right)+C$
Where $C$ is constant of integration.