Question

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

Answer 1

$\frac{1}{(x^2+1)(x^2+4)}=\frac{Ax+B}{(x^2+1)}+\frac{Cx+D}{(x^2+4)}$

$\Rightarrow 1=(Ax+B)(x^2+4)+(Cx+D)(x^2+1)$

$\Rightarrow 1=A{x}^3+4Ax+B{x}^2+C{x}^3+Cx+D{x}^2+D$

Equating the coefficients of $x^3,x^2,x$, and constant term, we obtain
$A+C=0$

$B+D=0$

$4A+C=0$

$4B+D=1$

On solving these equations, we obtain

$A=0,B=\frac{1}{3},C=0$, and $D=-\frac{1}{3}$

From equation (1), we obtain

$\frac{1}{(x^2+1)(x^2+4)}=\frac{1}{3(x^2+1)}-\frac{1}{3(x^2+4)}$

$\therefore \int \frac{1}{(x^2+1)(x^2+4)}dx=\frac{1}{3}\int \frac{1}{x^2+1}dx-\frac{1}{3}\int \frac{1}{x^2+4}dx$

$=\frac{1}{3}{\tan }^{-1}x-\frac{1}{3}\cdot \frac{1}{2}{\tan }^{-1}\frac{x}{2}+C$

$=\frac{1}{3}{\tan }^{-1}x-\frac{1}{6}{\tan }^{-1}\frac{x}{2}+C$