Question

$\underset{0}{\overset{1}{\int }}{\cot }^{-1}\left(1-x+x^2\right)dx$

Answer 1

$$\begin{gathered} {\int }_0^{1}co{t}^{-1}\left(1-x+x^2\right)dx \\ ={\int }_0^{1}co{t}^{-1}\left[x\left(x-1\right)+1\right]dx \\ ={\int }_0^{1}co{t}^{-1}\left[\frac{\left(x\left(x-1\right)+1\right)}{x-\left(x-1\right)}\right]dx \\ ={\int }_0^{1}co{t}^{-1}x-co{t}^{-1}\left(x-1\right)dx \\ ={\left[xco{t}^{-1}x\right]}_0^1+{\int }_0^1\frac{x}{1+x^2}dx-{\left[\left(x-1\right)co{t}^{-1}\left(x-1\right)\right]}_0^1-{\int }_0^1\frac{\left(x-1\right)}{1+{\left(x-1\right)}^2}dx \\ ={\left[xco{t}^{-1}x\right]}_0^1+\frac{1}{2}{\left[\log \left(1+x^2\right)\right]}_0^1-{\left[\left(x-1\right)co{t}^{-1}\left(x-1\right)\right]}_0^1-\frac{1}{2}{\left[\log \left(1+{\left(1-x\right)}^2\right)\right]}_0^1 \\ =\frac{\mathrm{\pi }}{4}-\frac{1}{2}\log 2+\frac{\mathrm{\pi }}{4}-\frac{1}{2}\log 2 \\ =\frac{\mathrm{\pi }}{2}-\log 2 \end{gathered}$$