Question

The value of ${\int }_0^1{\cot }^{-1}(1+x^2-x)dx$ is

• A. $\pi /2-log2$
• B. $\pi -log2$
• C. $\pi /4-log2$
• D. $2{\int }_0^1{\tan }^{-1}x:dx$

Answer 1

Answer: (A), (D)

The correct options are
A $\pi /2-log2$
D $2{\int }_0^1{\tan }^{-1}x:dx$

${\cot }^{-1}(1+x^2-x)={\tan }^{-1}\left(\frac{x+1-x}{1-x(1-x)}\right)={\tan }^{-1}x+{\tan }^{-1}(1-x)$
$I={\int }_0^1{\cot }^{-1}(1+x^2-x)dx={\int }_0^1{\tan }^{-1}xdx+{\int }_0^1{\tan }^{-1}(1-x)dx$
${\int }_0^1{\tan }^{-1}xdx+{\int }_0^1{\tan }^{-1}xdx=2{\int }_0^1{\tan }^{-1}xdx$
${=2x{\tan }^{-1}x]}_0^1-{\int }_0^1\frac{2x}{1+x^2}dx$
$={2{\tan }^{-1}\left(1\right)-\log (1+x^2)]}_0^1$
$=2\left(\pi /4\right)-\log 2=\pi /2-\log 2$