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 $0$

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

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

$={\int }_0^1{\tan }^{-1}xdx+{\int }_0^1{\tan }^{-1}(x-1)dx$

Using the formula for ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

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

$=-{\int }_0^1{\tan }^{-1}(x-1)dx+{\int }_0^1{\tan }^{-1}(x-1)dx=0$