Question

Evaluate: ${\int }_0^1\frac{x^3}{1+x^8}dx$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{8}$
• C. $\frac{\pi }{16}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $\frac{\pi }{16}$

Let, $z=x^4⟹dz=4x^{3}dx⟹x^{3}dx=\frac{dz}{4}$

When $x=0,z=0$ and when $x=1,z=1$

Hence,integration becomes:-

$\frac{1}{4}{\int }_0^1\frac{dz}{1+z^2}$

$=\frac{1}{4}{\left[{\tan }^{-1}z\right]}_0^1$

$=\frac{1}{4}[\frac{\pi }{4}-0]$

$=\frac{\pi }{16}$

Hence, answer is option-(C).