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

Answer: (B)

$0$

Let $={\int }_0^1{\tan }^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$
$={\int }_0^1{\tan }^{-1}\left[\frac{x(x-1)}{1-x(x-1)}\right]dx$
${\int }_0^1\{{\tan }^{-1}x+{\tan }^{-1}(x-1)\}dx[\because {\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)]$
Also
$I={\int }_0^1\{{\tan }^{-1}(x-1)-{\tan }^{-1}1-(x-1)\}dx[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx]$
$I={\int }_0^1\{{\tan }^{-1}(x-1)-{\tan }^{-1}1-\left(x\right)\}$
On adding Eq.(i) and (ii), we get
$2I=0$

$\Rightarrow I=0$