Question

integral of 
$\frac{x^3+1}{(x^4+1)(x+1)}$

Answer 1

Well this is done by factorization first. and then integration.

$\frac{x^3+1}{(x^4+1)(1+x)}=\frac{(1+x)(x^2-x+1)}{(1+x)(1+x^4)}=\frac{x^2-x+1}{(x^2+1)^2-2x^2}\\\\=\frac{x^2+1-\sqrt2 x+(\sqrt2-1)x}{(x^2+1-\sqrt2x)(x^2+1+\sqrt2 x)}=\frac{1}{x^2+\sqrt2 x +1}+\frac{(\sqrt2 -1)x}{x^4+1}\\\\=\frac{1}{(x+\frac{1}{\sqrt2})^2+(\frac{1}{\sqrt2})^2}+\frac{\sqrt2-1}{2}\frac{2x}{1+(x^2)^2}\\\\Integral=\sqrt2\ Tan^{-1}(\sqrt2x+1)+\frac{\sqrt2-1}{2}Tan^{-1}(x^2)+K$