Question

Evaluate $\int x^2{\tan }^{-1}xdx$
(where $C$ is constant of integration)

• A. $-\frac{x^3}{3}{\tan }^{-1}x+\frac{x^2}{6}-\frac{1}{6}\ln \mid x^2+1\mid +C$
• B. $\frac{x^3}{3}{\tan }^{-1}x+\frac{x^2}{6}+\frac{1}{6}\ln \mid x^2+1\mid +C$
• C. $\frac{x^3}{3}{\tan }^{-1}x-\frac{x^2}{6}-\frac{1}{6}\ln \mid x^2+1\mid +C$
• D. $\frac{x^3}{3}{\tan }^{-1}x-\frac{x^2}{6}+\frac{1}{6}\ln |x^2+1|+C$

Answer 1

The correct option is D $\frac{x^3}{3}{\tan }^{-1}x-\frac{x^2}{6}+\frac{1}{6}\ln |x^2+1|+C$

Let $I=\int x^2{\tan }^{-1}xdx$
Using ILATE rule, we get
$I=\left({\tan }^{-1}x\right)\frac{x^3}{3}-\int \frac{x^3}{3}\cdot \frac{dx}{x^2+1}$
Put $x^2=t$
$\Rightarrow 2xdx=dt$
$=\frac{x^3}{3}{\tan }^{-1}x-\frac{1}{3}\int \frac{t\cdot dt}{2(t+1)}=\frac{x^3}{3}{\tan }^{-1}x-\frac{1}{6}\int \frac{t+1-1}{t+1}dt=\frac{x^3}{3}{\tan }^{-1}x-\frac{1}{6}(t-\ln |t+1|)=\frac{x^3}{3}{\tan }^{-1}x-\frac{x^2}{6}+\frac{1}{6}\ln \mid x^2+1\mid +C$