Question

If $2{\int }_0^1{\tan }^{-1}xdx={\int }_0^1{\cot }^{-1}(1-x+x^2)dx$, then ${\int }_0^1{\tan }^{-1}(1-x+x^2)dx$ is equal to:

• A. $\log 2$
• B. $\frac{\pi }{2}-\log 4$
• C. $\frac{\pi }{2}+\log 2$
• D. $\log 4$

Answer 1

The correct option is A $\log 2$

We know that ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$
$2{\int }_0^1{\tan }^{-1}xdx={\int }_0^1{\cot }^{-1}(1-x+x^2)dx$,
then $2{\int }_0^1{\tan }^{-1}xdx={\int }_0^1[\frac{\pi }{2}-{\tan }^{-1}(1-x+x^2\left)\right]dx$
$I_f={\int }_0^1{\tan }^{-1}(1-x+x^2)dx=\frac{\pi }{2}-2{\int }_0^1{\tan }^{-1}xdx$
Integrating ${\int }_0^1{\tan }^{-1}xdx$
$I={\int }_0^1{\tan }^{-1}xdx$
$\left[\right({\tan }^{-1}x)x{]}_0^1-{\int }_0^1\frac{x}{1+x^2}$
$=\frac{\pi }{4}-{\int }_0^1\frac{x}{1+x^2}$
$=\frac{\pi }{4}-\frac{1}{2}ln2$
$I_f=\frac{\pi }{2}-2[\frac{\pi }{4}-\frac{1}{2}ln2]$
$I_f=ln2$