Question

The value of the integral ${\int }_0^1\frac{x^3}{1+x^3}dx$ is equal to

• A. $\frac{\pi }{8}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{16}$
• D. $\frac{\pi }{6}$
• E. $\frac{\pi }{12}$

Answer 1

Answer: (C)

$\frac{\pi }{16}$

Let $I={\int }_0^1\frac{x^3}{1+x^3}dx$
Put $x^4=t\Rightarrow 4x^{3}dx=dt$
Therefore, $I={\int }_0^1\frac{1}{1+t^2}.\frac{1}{4}dt=\frac{1}{4}[{\tan }^{-1}t{]}_0^1$
$=\frac{1}{4}[{\tan }^{-1}1-{\tan }^{-1}0]=\frac{1}{4}[\frac{\pi }{4}-0]=\frac{\pi }{16}$