Question

Evaluate : $\int \frac{x^2}{(x^2+4)(x^2+9)}dx$.

Answer 1

$I=\int \frac{x^2}{(x^2+4)(x^2+9)}dx$

$\Rightarrow \int \frac{x^2}{(x^2+4)(x^2+9)}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{x^2+9}$

$=\frac{(x^2+4)(Cx+D)+(x^2+9)(Ax+B)}{(x^2+4)(x^2+9)}$

Denominators are equal, so we require equality of numerators:
$x^2=x^3(A+C)+x^2(B+D)+x(9A+4C)+9B+4D$

Coefficients near like terms should be equal, so the following system is obtained:
$A+C=0,B+D=1,9A+4C=0,9B+4D=0$

By solving it, we get that

$A=0,B=\frac{-4}{5},C=0,D=\frac{9}{5}$

Therefore, $I=\frac{-4}{5(x^2+4)}+\frac{9}{5(x^2+9)}dx$

Integrate term by term,

$=\frac{-4}{5}\int \frac{1}{x^2+4}dx+\frac{9}{5}\int \frac{1}{x^2+9}$

We know that $\int \frac{1}{x^2+1}dx={\tan }^{-1}x$

So, $I=\frac{-4}{5}.\frac{1}{2}{\tan }^{-1}\frac{x}{2}+\frac{9}{5}.\frac{1}{3}{\tan }^{-1}\frac{x}{3}+C$

$I=\frac{3}{5}{\tan }^{-1}\left(\frac{x}{3}\right)-\frac{2}{5}{\tan }^{-1}\left(\frac{x}{2}\right)+C$