Question

The value of definite integral $\int\limits_1^{\sqrt{2}} x \tan ^{-1}\left(x^2-1\right) d x$ 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} \int\limits_0^1 \tan ^{-1} tdt =\frac{1}{2}\left[\left.\tan ^{-1} t \cdot t \right|_0 ^1-\int\limits_0^1 \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} $