Question

Sol. [Key Concept: Use $\cot^{- 1}(1 + x + x^{2}) = \tan^{- 1}(x + 1) - \tan^{- 1}x$, then integrate by parts.]

$$= \int_{0}^{1}\mspace{2mu}\tan^{- 1}\left( \frac{1}{1 + x + x^{2}} \right)dx$$

$$\begin{matrix} = & \ \int_{0}^{1}\mspace{2mu}\mspace{2mu}\tan^{- 1}\left( \frac{(x + 1) - x}{1 + x(x + 1)} \right)dx \\ = & \ \int_{0}^{1}\mspace{2mu}\mspace{2mu}\left( \tan^{- 1}(x + 1) - \tan^{- 1}x \right)dx \\ = & \left( x\tan^{- 1}(x + 1) - \frac{1}{2}ln\left| 1 + (1 + x)^{2} \right| + \tan^{- 1}(x + 1) \right)_{0}^{1} \\ & \ - \left( x\tan^{- 1}x - \frac{1}{2}ln\left| 1 + x^{2} \right| \right)_{0}^{1} \\ = & \left( 2\tan^{- 1}2 - \frac{\pi}{4} - \frac{1}{2}ln\frac{5}{2} \right) - \left( \frac{\pi}{4} - \frac{1}{2}ln2 \right) \\ = & 2\tan^{- 1}2 - \frac{1}{2}ln\left( \frac{5}{4} \right) - \frac{\pi}{2} \end{matrix}$$