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. None of these

Answer 1

The correct option is B

$0$

The explanation for the correct answer.

Solve for the given integral.

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

Hence, option(B) is the correct answer.