Question

${\int }_0^1\frac{x^2}{1+x^6}dx$

• A. $\frac{\pi }{12}$
• B. $\frac{\pi }{8}$
• C. $\pi$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A $\frac{\pi }{12}$

${\int }_0^1\frac{x^2}{1+x^6}dx$
$I={\int }_0^1\frac{x^2}{1+(x^3{)}^2}dx$
$I=\frac{1}{3}{\int }_0^1\frac{1}{1+t^2}dt$
So, $I=\frac{1}{3}{\left[{\tan }^{-1}t\right]}_0^1$
$\Rightarrow I=\frac{\pi }{12}$