Question

Evaluate : $\int \frac{x^2+1}{(x^2+4)(x^2+25)}dx$.

Answer 1

$I=\int \frac{x^2+1}{(x^2+4)(x^2+25)}dx$

$=\int \frac{x^2+1}{(x^2+4)(x^2+25)}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{x^2+25}$

$=\frac{(x^2+4)(Cx+D)+(x^2+25)(Ax+B)}{(x^2+4)(x^2+25)}$

Denominators are equal, so we require equality of numerators:
$x^2+1=x^3(A+C)+x^2(B+D)+x(25A+4C)+25B+4D$

Coefficients near like terms should be equal, so the following system is obtained:
$A+C=0,B+D=1,25A+4C=0,25B+4D=0$

By solving it, we get that

$A=0,B=\frac{-1}{7},C=0,D=\frac{8}{7}$

Therefore, $I=\frac{-1}{7(x^2+4)}+\frac{8}{7(x^2+25)}dx$

Integrate term by term,
$=\frac{-1}{7}\int \frac{1}{x^2+4}dx+\frac{8}{7}\int \frac{1}{x^2+25}$

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

So, $I=\frac{-1}{7}.\frac{1}{2}{\tan }^{-1}\frac{x}{2}+\frac{8}{7}.\frac{1}{5}{\tan }^{-1}\frac{x}{5}+C$

$I=\frac{8}{35}{\tan }^{-1}\left(\frac{x}{5}\right)-\frac{1}{14}{\tan }^{-1}\left(\frac{x}{2}\right)+C$