Answer:
Option B
Explanation:
Ler l=$\int_{0}^{1}x \tan ^{-1} x dx$
=$[\tan^{-1}x\int x dx]^{1}_{0}-\int_{0}^{1} (\frac{d}{dx}\tan^{-1} x \int x dx) dx$
=$\left[ \tan ^{-1} .x \frac{x^2}{2}\right]^{1}_{0}-\int_{0}^{1}\left(\frac{1}{1+x^{2}}.\frac{x^{2}}{2}\right)dx$
=$\left(\frac{1}{2} \tan^{-1}1-0\right)-\frac{1}{2}\int_{0}^{1}\frac{1+x^{2}-1}{1+x^{2}} dx$
=$\frac{1}{2}\frac{\pi}{4}-\frac{1}{2}\int_{0}^{1} \left(1-\frac{1}{1+x^{2}}\right) dx$
=$\frac{\pi}{8}-\frac{1}{2}[x-tan^{-1}x]^{1}_{0}$
= $\frac{\pi}{8}-\frac{1}{2}[1-tan^{-1}1-0+0]$
= $\frac{\pi}{8}-\frac{1}{2}[1-\frac{\pi}{4}]=\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}=\frac{\pi}{4}-\frac{1}{2}$
