Question

${\int }_0^{1}x($ tan${}^{-1}x{)}^{2}dx=$

• A. $\frac{{\pi }^2}{6}-\frac{\pi }{3}+\frac{1}{2}$ log 2
• B. $\frac{{\pi }^2}{16}+\frac{\pi }{4}+\frac{1}{2}$ log2
• C. $\frac{{\pi }^2}{16}-\frac{\pi }{4}+\frac{1}{2}$ log2
• D. $\frac{{\pi }^2}{16}$

Answer 1

Answer: (C)

$\frac{{\pi }^2}{16}-\frac{\pi }{4}+\frac{1}{2}$ log2

${\int }_0^{1}x{\left({\tan }^{-1}x\right)}^{2}dx$
$={\left[\frac{x^2}{2}{\left({\tan }^{-1}x\right)}^2\right]}_0^1-\int \frac{x^2}{2}2.{\tan }^{-1}x\frac{1}{1+x^2}dx$
$=\frac{{\pi }^2}{32}-{\left[{\tan }^{-1}x(x-{\tan }^{-1}x)\right]}_0^1+{\int }_0^1\frac{1}{1+x^2}(x-{\tan }^{-1}x)dx$
$=\frac{{\pi }^2}{3^2}-\frac{\pi }{4}+\frac{{\pi }^2}{16}+{\int }_0^1\frac{x}{1+x^2}-\int \frac{{\tan }^{-1}x}{1+x^2}dx$
$=\frac{{\pi }^2}{3^2}-\frac{\pi }{4}+\frac{{\pi }^2}{16}+{\left[\frac{1}{2}\log (1+x^2)\right]}_0$
$=\frac{{\pi }^2}{16}-\frac{\pi }{4}+\frac{1}{2}\log 2$