Answer will be **1/2**, not 1/3.

$\displaystyle\int_{0}^{1}x\tan^{-1}\left(\frac{1}{x}\right)dx$

$=\displaystyle\left[\tan^{-1}\left(\frac{1}{x}\right)\cdot\frac{x^{2}}{2}-\frac{1}{2} \int \frac{x^{4}}{x^{2}+1}\color{blue}{\left(\frac{-1}{x^2}\right)}dx \right]_{0}^{1}$

$=\displaystyle\left[\tan^{-1}\left(\frac{1}{x}\right)\cdot\frac{x^{2}}{2}\right]_0^1+\frac{1}{2}\left[ \int \frac{x^2}{x^{2}+1}dx \right]_{0}^{1}$

$\displaystyle=\frac{\Pi }{8}+\frac{1}{2}\left[ \int \frac{(x^2+1)-1}{x^{2}+1}dx \right]_{0}^{1}$

$\displaystyle=\frac{\Pi}{8}+ \frac{1}{2}\left[x-tan^{-1}x \right]_{0}^{1}$

$=\frac{\Pi}{8}+\frac{1}{2}-\frac{\Pi}{8}$

$=1/2$

Hence, answer should be $1/2.$