Question

Integrate the following functions w.r.t. x.

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

Answer 1

$Let\frac{1}{(x^2+1)(x^2+4)}dx=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+4}\Rightarrow 1=(Ax+B)(x^2+4)+(Cx+D)(x^2+1)\text{On comparing the coefficients of}{x}^3\text{,}{x}^2\text{,x and constant term on both sides, we get}A+C=0,B+D=0,4A+C=0and4B+D=1\text{On solving these equations, we get}A=0,C=0,B=\frac{1}{3}andD=-\frac{1}{3}\therefore \int \frac{1}{(x^2+1)(x^2+4)}dx=\frac{1}{3}\int (\frac{1}{x^2+1}-\frac{1}{(x^2+4)})dx=\frac{1}{3}\{ta{n}^{-1}x-\frac{1}{2}ta{n}^{-1}\left(\frac{x}{2}\right)\}+C$