Question

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

• A. $\frac{\pi }{2}+ln2$
• B. $\frac{\pi }{2}-ln2$
• C. $\pi -ln2$
• D. None of these

Answer 1

The correct option is B $\frac{\pi }{2}-ln2$

$\begin{array}{c}{\int }_0^1{\cot }^{-1}\left(1-x+x^2\right)dx={\int }_0^1{\tan }^{-1}\left(\frac{1}{1-x+x^2}\right)dx\\ {\int }_0^1{\tan }^{-1}\left(\frac{x-\left(x-1\right)}{1+x\left(x-1\right)}\right)dx={\int }_0^1{\tan }^{-1}x-{\int }_0^1{\tan }^{-1}\left(x-1\right)dx\left[let\left(x-1\right)=t\right]\\ ={\int }_0^1{\tan }^{-1}x-{\int }_{-1}^0{\tan }^{-1}tdt\\ ={{\left[x{\tan }^{-1}x-\frac{1}{2}\log \left(1+x^2\right)\right]}_0}^1-{{\left[t{\tan }^{-1}t-\frac{1}{2}\log \left(1+t^2\right)\right]}^0}_{-1}\\ =\left(\frac{\pi }{4}-\frac{1}{2}\log 2\right)-\left[0-\left(\frac{\pi }{4}-\frac{1}{2}\log 2\right)\right]\\ =\left(\frac{\pi }{4}-\frac{1}{2}\log 2\right)\times 2\\ =\frac{\pi }{2}-\log 2\end{array}$

Hence,

Option $B$ is correct answer.