Question

Evaluate ${\int }_0^1{\cot }^{-1}(1-x+x^2)dx$

Answer 1

$\begin{array}{c}{\int }_0^1{\cot }^{-1}\left(1-x+x^2\right)dx\Rightarrow {\int }_0^1\frac{1}{1-x+x^2}dx\\ {\int }_0^1{\tan }^{-1}\frac{1}{1-x\left(1-x\right)}dx\Rightarrow {\int }_0^1{\tan }^{-1}\frac{x+\left(1-x\right)}{1-x\left(1-x\right)}\left[{\tan }^{-1}\left(\frac{a+b}{1-ab}\right)={\tan }^{-1}a+{\tan }^{-1}b\right]\\ {\int }_0^1{\tan }^{-1}xdx+{\int }_0^1{\tan }^{-1}\left(1-x\right)dx\left[{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b+x\right)dx\right]\\ {\int }_0^1{\tan }^{-1}xdx+{\int }_0^1{\tan }^{-1}xdx\\ 2{\int }_0^{1}1{\tan }^{-1}xdx\left[un\sin gbyparts\right]\\ 2\left[{\tan }^{-1}x\cdot x-\int \frac{x}{1+x^2}dx\right]\left[\begin{array}{c}let1+x^2=t\\ difference=\\ 2xdx=dt\\ xdx=\frac{dt}{2}\end{array}\right]\\ 2\left[x\cdot {\tan }^{-1}x-\frac{1}{2}\int \frac{dt}{t}\right]\\ 2\left[x{\tan }^{-1}x-\frac{1}{2}\log t\right]\\ 2{\left[x{\tan }^{-1}x-\frac{1}{2}\log \left(1+x^2\right)\right]}_0^1\\ 2\left[\frac{\pi }{4}-\frac{1}{2}\log \sqrt{2}\right]\\ =\frac{\pi }{2}-\log \sqrt{2}\end{array}$