Question

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

Answer 1

$\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=1-\frac{(4x^2+10)}{(x^2+3)(x^2+4)}$
Let $\frac{4x^2+10}{(x^2+3)(x^2+4)}=\frac{Ax+B}{(x^2+3)}+\frac{Cx+D}{(x^2+4)}$
$\Rightarrow 4x^2+10=(Ax+B)(x^2+4)+(Cx+D)(x^2+3)$
$4x^2+10=A{x}^3+4Ax+B{x}^2+4B+C{x}^3+3Cx+D{x}^2+3D$
$4x^2+10=(A+C)x^3+(B+D)x^2+(4A+3C)x+(4B+3D)$
Equating the coefficients of $x^3,x^2,x$, and constant term, we obtain
$A+C=0$
$B+D=4$
$4A+3C=0$
$4B+3D=10$
On solving these equations, we obtain
$A=0,B=2,C=0$, and $D=6$
$\therefore \frac{4x^2+10}{(x^2+3)(x^2+4)}=\frac{-2}{(x^2+3)}+\frac{6}{(x^2+4)}$
$\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=1-(\frac{-2}{(x^2+3)}+\frac{6}{(x^2+4)})$
$\Rightarrow \int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx=\int \{1+\frac{2}{(x^2+3)}-\frac{6}{(x^2+4)}\}dx$
$=\int \{1+\frac{2}{x^2+(\sqrt{3}{)}^2}-\frac{6}{x^2+2^2}\}$
$=x+2\left(\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{x}{\sqrt{3}}\right)-6\left(\frac{1}{2}{\tan }^{-1}\frac{x}{2}\right)+C$
$=x+\frac{2}{\sqrt{3}}{\tan }^{-1}\frac{x}{\sqrt{3}}-3{\tan }^{-1}\frac{x}{2}+C$