Question

The value of definite integral $\underset{1}{\overset{\sqrt{2}}{\int }}x{\tan }^{-1}\left(x^2-1\right)dx$ equals

• A. $\frac{\pi }{4}-\frac{\ln 3}{2}$
• B. $\frac{\pi }{8}-\frac{\ln 2}{4}$
• C. $\frac{\pi }{6}-\frac{\ln 2}{8}$
• D. $\frac{\pi }{2}-\frac{\ln 5}{4}$

Answer 1

Answer: (B)

Put $x^2-1=t$
$I=\frac{1}{2}\underset{0}{\overset{1}{\int }}{\tan }^{-1}tdt=\frac{1}{2}\left[{{\tan }^{-1}t\cdot t|}_0^1-\underset{0}{\overset{1}{\int }}\frac{t}{1+t^2}dt\right]$
$I=\frac{\pi }{8}-\frac{1}{2}\cdot \frac{1}{2}\ln (2)=\frac{\pi }{8}-\frac{\ln 2}{4}$