Question

The integral ${\int }_0^1\frac{x^3}{1+x^8}dx=$

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

Answer 1

Answer: (A)

$\frac{\pi }{16}$

Let $I={\int }_0^1\frac{x^3}{1+x^8}dx$

Now, $\int \frac{x^3}{1+x^8}dx=\frac{1}{4}\int \frac{d\left(x^4\right)}{1+x^8}=\frac{1}{4}{\tan }^{-1}\left(x^4\right)+c$

$I={\int }_0^1\frac{x^3}{1+x^8}dx={\left(\frac{1}{4}{\tan }^{-1}\right(x^4)+c)}_0^1$

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

$=\frac{1}{4}{\tan }^{-1}\left(1\right)$

$=\frac{1}{4}\cdot \frac{\pi }{4}=\frac{\pi }{16}$

Hence, ${\int }_0^1\frac{x^3}{1+x^8}dx=\frac{\pi }{16}$