Question

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

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

Answer 1

Answer: (D)

$\frac{\pi }{4}-\frac{1}{2}\log 2.$

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