Evaluate: ∫_0^1 x tan^-1 x d x.

$\displaystyle \int_0^1 x\tan^{-1}x\,dx = \left[\tan^{-1}x \times \dfrac{x^2}{2}\right]_0^1 - \int_0^1 \dfrac{1}{1+x^2}\times\dfrac{x^2}{2}\,dx$

$\displaystyle = \dfrac{\pi}{8} - \int_0^1 \dfrac{1+x^2-1}{1+x^2}\times\dfrac{1}{2}\,dx$

$\displaystyle = \dfrac{\pi}{8} - \dfrac{1}{2}\left[\int_0^1 dx - \int_0^1 \dfrac{1}{1+x^2}\,dx\right]$

$\displaystyle = \dfrac{\pi}{8} - \dfrac{1}{2}\left[x-\tan^{-1}x\right]_0^1$

$\displaystyle = \dfrac{\pi}{8} - \dfrac{1}{2} + \dfrac{\pi}{8} = \dfrac{\pi}{4} - \dfrac{1}{2}$