Question

Evaluate the integral
${\int }_0^1{\tan }^{-1}xdx$

• A. $\frac{\pi }{4}-\frac{1}{2}log2$
• B. $\frac{\pi }{4}-\frac{1}{4}log2$
• C. $\frac{\pi }{4}+\frac{1}{2}log2$
• D. $\frac{\pi }{4}+\frac{1}{4}log2$

Answer 1

The correct option is A $\frac{\pi }{4}-\frac{1}{2}log2$

${\int }_0^1{\tan }^{-1}x.dx$

$\int {\tan }^{-1}x\cdot dx=x.{\tan }^{-1}x-\int \frac{1}{1+x^2}\cdot x.dx$

$=x{\tan }^{-1}x-\frac{1}{2}\int \frac{2xdx}{1+x^2}$

$=x{\tan }^{-1}x-\frac{1}{2}\left[\log \right(1+x^2\left)\right]$

$=x{\tan }^{-1}x-\frac{\log (1+x^2)}{2}$

${\int }_0^1{\tan }^{-1}x.dx=x.{\tan }^{-1}x-\frac{\log (1+x^2)}{2}{|}_0^1$

$=[1.{\tan }^{-1}1-\frac{\left(2\right)}{2}]-[0-\log (1\left)\right]$

$={\tan }^{-1}1-\frac{1}{2}\log 2=\frac{\pi }{4}-\frac{1}{2}\log 2$