Question

The value of ${\int }_0^1{\tan }^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$ is

• A. $1$
• B. $0$
• C. $-1$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is B $0$

Let $I={\int }_0^1{\tan }^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$
$\Rightarrow I={\int }_0^1{\tan }^{-1}\left(\frac{x-(1-x)}{1+x(1-x)}\right)dx$
$\Rightarrow I={\int }_0^1[{\tan }^{-1}x-{\tan }^{-1}(1-x\left)\right]dx$ ................ (1)
$\Rightarrow I={\int }_0^1\left[{\tan }^{-1}\right(1-x)-{\tan }^{-1}(1-1+x\left)\right]dx$
$\Rightarrow I={\int }_0^1\left[{\tan }^{-1}\right(1-x)-{\tan }^{-1}(x\left)\right]dx$
$\Rightarrow I={\int }_0^1\left[{\tan }^{-1}\right(1-x)-{\tan }^{-1}(x\left)\right]dx$ ........... (2)
Adding (1) and (2), we obtain
$2I={\int }_0^1({\tan }^{-1}x+{\tan }^{-1}(1-x)-{\tan }^{-1}(1-x)-{\tan }^{-1}x)dx=0$
$\Rightarrow I=0$
Hence, the correct Answer is B.