The integral $\int x ^ { 2 } \tan ^ { - 1 } x d x$ is

A) $- \frac { 2 x ^ { 3 } \tan ^ { - 1 } x + \left( 1 + x ^ { 2 } \right) - \ln \left( 1 + x ^ { 2 } \right) } { 6 } + C$
B) $- \frac { 2 x ^ { 3 } \tan ^ { - 1 } x - \left( 1 + x ^ { 2 } \right) + \ln \left( 1 + x ^ { 2 } \right) } { 6 } + C$
C) $\frac { 2 x ^ { 3 } \tan ^ { - 1 } x + \left( 1 + x ^ { 2 } \right) - \ln \left( 1 + x ^ { 2 } \right) } { 6 } + C$
D) $\frac { 2 x ^ { 3 } \tan ^ { - 1 } x - \left( 1 + x ^ { 2 } \right) + \ln \left( 1 + x ^ { 2 } \right) } { 6 } + C$
E) $\frac { 2 x ^ { 3 } \tan ^ { - 1 } x + \left( 1 + x ^ { 2 } \right) + \ln \left( 1 + x ^ { 2 } \right) } { 6 } + C$

Question

The integral $\int x ^ { 2 } \tan ^ { - 1 } x d x$ is

A) $- \frac { 2 x ^ { 3 } \tan ^ { - 1 } x + \left( 1 + x ^ { 2 } \right) - \ln \left( 1 + x ^ { 2 } \right) } { 6 } + C$
B) $- \frac { 2 x ^ { 3 } \tan ^ { - 1 } x - \left( 1 + x ^ { 2 } \right) + \ln \left( 1 + x ^ { 2 } \right) } { 6 } + C$
C) $\frac { 2 x ^ { 3 } \tan ^ { - 1 } x + \left( 1 + x ^ { 2 } \right) - \ln \left( 1 + x ^ { 2 } \right) } { 6 } + C$
D) $\frac { 2 x ^ { 3 } \tan ^ { - 1 } x - \left( 1 + x ^ { 2 } \right) + \ln \left( 1 + x ^ { 2 } \right) } { 6 } + C$
E) $\frac { 2 x ^ { 3 } \tan ^ { - 1 } x + \left( 1 + x ^ { 2 } \right) + \ln \left( 1 + x ^ { 2 } \right) } { 6 } + C$

Quizplus · Accepted Answer

This integral can be solved using integration by parts, where $u = \tan^{-1}x$ and $dv = x^2 dx$. The derivative of $\tan^{-1}x$ is $\frac{1}{1+x^2}$, and the integral of $x^2$ is $\frac{x^3}{3}$. Applying integration by parts formula and simplifying leads to the expression given in option D.

The Integral ∫x2tan⁡−1xdx\int x ^ { 2 } \tan ^ { - 1 } x d x∫x2tan−1xdx

The Integral $\int x ^ { 2 } \tan ^ { - 1 } x d x$