Question

Evaluate $\int \left(\frac{x^6-1}{1+x^2}\right)dx$ for $x\in R.$

Answer 1

Let $I=\int \left(\frac{x^6-1}{x^2+1}\right)dx$

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

By polynomial division

$\Rightarrow I=\int [\frac{-2}{x^2+1}+x^4-x^2+1]dx$

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

$\Rightarrow I=-2{\tan }^{-1}\left(x\right)+\frac{x^5}{5}-\frac{x^3}{3}+x+c$