Question

The value of $\underset{0}{\overset{1}{\int }}\frac{x^{-5/8}}{1+x^{1/4}}dx$ is

• A. $\frac{1}{8}(1-\frac{\pi }{4})$
• B. $1-\frac{\pi }{4}$
• C. $4-\pi$
• D. $8-2\pi$

Answer 1

The correct option is D $8-2\pi$

Let $t=x^{1/8}dt=\frac{1}{8}{x}^{-7/8}dxdx=8x^{7/8}dt$

$I=\underset{0}{\overset{1}{\int }}\frac{8x^{-5/8}.x^{7/8}}{1+t^2}dt$
$=8\underset{0}{\overset{1}{\int }}\frac{t^2}{1+t^2}dt$
$=8\underset{0}{\overset{1}{\int }}[1-\frac{1}{1+t^2}]dt$
$=8{[t-{\tan }^{-1}t]}_0^1$
$=8(1-\frac{\pi }{4})=8-2\pi$