Question

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

Answer 1

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

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

$I=\frac{-1}{3}ta{n}^{-1}x+\frac{4}{3}\frac{1}{2}ta{n}^{-1}\frac{x}{2}+c$

$I=\frac{2}{3}ta{n}^{-1}\frac{x}{2}-\frac{1}{3}ta{n}^{-1}x+c$