$\int {(\frac{x}{1 + x^{5}})}^{3 / 2} d x$ equals to

\frac{2}{5} \sqrt{\frac{x^{5}}{1 + x^{5}}} + c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{2}{5} \sqrt{\frac{x}{1 + x^{5}}} + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2}{5} \sqrt{\frac{1}{1 + x^{5}}} + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{2}{5} \sqrt{\frac{x^{5}}{1 + x^{5}}} + c$
$I = \int {(\frac{x}{(1 + x^{5})})}^{3 / 2} d x = \int \frac{x^{- 6}}{(1 + x^{- 5})^{3 / 2}} d x$

Substitute $t = 1 + x^{- 5}; d t = - 5 x^{- 6} d x$

$∴ I = - \frac{1}{5} \int \frac{d t}{(t)^{3 / 2}} = \frac{2}{5} t^{- 1 / 2} + c$

$= \frac{2}{5} (1 + x^{- 5})^{- 1 / 2} + c$

$= \frac{2}{5} \sqrt{\frac{x^{5}}{1 + x^{5}}} + c$

Suggest Corrections

Similar questions

\int (x + x^{2} + x^{3} + x^{4})

\int x^{13 / 2} {(1 + x^{5 / 2})}^{1 / 2} d x

\frac{2}{5} {(1 + x^{5 / 2})}^{3 / 2} [\frac{2}{7} {(1 + x^{5 / 2})}^{2} - a (1 + x^{b}) + \frac{2}{3}] + c

lim x \to \infty \frac{\sqrt[7]{x^{7} - 1} \sqrt[5]{x^{5} + 2} \sqrt[9]{x^{9} - 2}}{\sqrt[6]{x^{6} + 1} \sqrt[5]{x^{5} + 1} \sqrt[4]{x^{4} + 4}} =

\frac{x}{5} = 1

x = \frac{1}{5}

x + \frac{1}{x} = 2

x^{5} + \frac{1}{x^{5}}

Join BYJU'S Learning Program