Question

The value of $\underset{0}{\overset{1}{\int }}{\tan }^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx,$ is

(a) 1
(b) 0
(c) −1
(d) π/4

Answer 1

(b) 0

$$\begin{gathered} \mathrm{Let},I={\int }_0^1{\tan }^{-1}\frac{2x-1}{1+x-x^2}dx...\left(\mathrm{i}\right) \\ ={\int }_0^1{\tan }^{-1}\frac{2\left(1-x\right)-1}{1+\left(1-x\right)-{\left(1-x\right)}^2}dx \\ ={\int }_0^1{\tan }^{-1}\frac{1-2x}{2-x-1-x^2+2x}dx \\ ={\int }_0^1{\tan }^{-1}\frac{1-2x}{1+x-x^2}dx \\ =-{\int }_0^1{\tan }^{-1}\frac{2x-1}{1+x-x^2}dx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ 2I={\int }_0^1{\tan }^{-1}\frac{2x-1}{1+x-x^2}dx-{\int }_0^1{\tan }^{-1}\frac{2x-1}{1+x-x^2}dx \\ =0 \\ \mathrm{Hence},I=0 \end{gathered}$$