∫x^2tan^-1x dx is :

The correct answer is Option (1) → $\frac{x^3}{3}tan^{-1}x-\frac{1}{6}x^2+\frac{1}{6}log|x^2+1|+C$

$∫x^2\tan^{-1}xdx$

$=\frac{x^3}{3}\tan^{-1}x-\int\frac{x^3}{3}\frac{d}{dx}(\tan^{-1}x)dx$

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

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

$=\frac{x^3}{3}\tan^{-1}x-\frac{x^2}{6}+\frac{x}{6}\log|x^2+1|+C$