Question

The integral ${\int }_0^1\frac{(ta{n}^{-1}x{)}^3}{1+x^2}dx=$

• A. $\frac{{\pi }^4}{64}$
• B. $\frac{{\pi }^4}{256}$
• C. $\frac{{\pi }^4}{1024}$
• D. $\frac{{\pi }^4}{512}$

Answer 1

The correct option is C $\frac{{\pi }^4}{1024}$

${\int }_0^1\frac{({\tan }^{-1}x{)}^3}{1+x^2}dx$

$\int \frac{({\tan }^{-1}x{)}^3}{1+x^2}dx=\int \left({\tan }^{-1}x{)}^{3}d\right({\tan }^{-1}x)$

$=\frac{({\tan }^{-1}x{)}^4}{4}+c$

${\int }_0^1\frac{({\tan }^{-1}x{)}^3}{1+x^2}\cdot dx={(\frac{({\tan }^{-1}x{)}^4}{4}+c)}_0^1$

$=[\frac{({\tan }^{-1}1{)}^4}{4}+c]-[\frac{({\tan }^{-1}0{)}^4}{4}+1]$

$=[\frac{{\left(\frac{\pi }{4}\right)}^4}{4}+c]-[0+c]$

$=\frac{{\pi }^4}{4^5}=\frac{{\pi }^4}{1024}$