Question

${\int }_0^1{\tan }^{-1}xdx=$

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

Answer 1

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

taking $u={\tan }^{-1}x,v=1$

integrating by parts we get $\int 1\times {\tan }^{-1}xdx={\tan }^{-1}x\int 1dx-\int (\frac{1}{1+x^2}\int 1dx)dx=x{\tan }^{-1}x-\int \frac{x}{1+x^2}dx=x{\tan }^{-1}x-\frac{1}{2}\int \frac{2x}{1+x^2}dx=x{\tan }^{-1}x-\frac{1}{2}\ln 1+x^2$

${\int }_0^1{\tan }^{-1}xdx={[x{\tan }^{-1}x-\frac{1}{2}\ln 1+x^2]}_0^1=[\frac{\pi }{4}-\frac{1}{2}\ln 2]-0=\frac{\pi }{4}-\frac{1}{2}\ln 2$

so A is the correct option