Question

${\int }_0^1{\tan }^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx=$

• A. $0$
• B. $1$
• C. $\pi$
• D. $2\pi$

Answer 1

Answer: (A)

$0$

Let:-

$I={\int }_0^1{\tan }^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$

$\Rightarrow I={\int }_0^1{\tan }^{-1}\left(\frac{2\left(1-x\right)-1}{1+\left(1-x\right)-{\left(1-x\right)}^2}\right)dx$

$\Rightarrow I={\int }_0^1{\tan }^{-1}\left[\frac{2-2x-1}{1+1-x-\left(1+x^2-2x\right)}\right]dx$

$\Rightarrow I={\int }_0^1{\tan }^{-1}\left[\frac{1-2x}{2-x-1-x^2+2x}\right]dx$

$\Rightarrow I={\int }_0^1{\tan }^{-1}\left[\frac{1-2x}{1+x-x^2}\right]dx$

$\Rightarrow I=-{\int }_0^1{\tan }^{-1}\left(\frac{2x-1}{1+x+x^2}\right)$

$\Rightarrow 2I=0$

$\Rightarrow I=0$