Question

Evaluate : $\int \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}dx$

Answer 1

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

Let $y=x^2$

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

Resolving into partial fraction

$\frac{(y+1)(y+4)}{(y+3)(y-3)}=$$\frac{1}{4(y+3)}+$$\frac{27}{4(y-5)}$

$\therefore I=$$\int \frac{1}{4(y+3)}dx+$$\int \frac{27}{4(y-5)}dx$

$I=\frac{4}{\sqrt{3}}{\tan }^{-1}\frac{x}{\sqrt{3}}+\frac{27}{8\sqrt{5}}\ln |\frac{x-\sqrt{5}}{x+\sqrt{5}}|+c$