-1-x-x2 d x is equal to

The correct option is B 2(1+x^1/3)^3/2+C∫√(1+√(x))/√(x^2)dx = ∫ x^ 2/3 ( 1+x^1/3 )^1/2 dx =∫ x^ 2/3(1+x^1/3 )^1/2dx 1+x^1/3=t^2 x^ 2/3dx=6 t dt =∫ 6 t^2dt =2t^ ...

152

You visited us 152 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Binomial Integral

∫√1+√x/√x2 d ...

Question

$\int \frac{\sqrt{1 + \sqrt[3]{x}}}{\sqrt[3]{x^{2}}} d x$ is equal to

(1 + x^{1 / 3})^{3 / 2} + C

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

- (1 + x^{1 / 3})^{3 / 2} + C

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

2 (1 + x^{1 / 3})^{3 / 2} + C

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

None of these

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

Open in App

Solution

The correct option is C $2 (1 + x^{1 / 3})^{3 / 2} + C$
$\int \frac{\sqrt{1 + \sqrt[3]{x}}}{\sqrt[3]{x^{2}}} d x = \int x^{- \frac{2}{3}} {(1 + x^{\frac{1}{3}})}^{1 / 2} d x$
$= \int x^{- 2 / 3} {(1 + x^{1 / 3})}^{1 / 2} d x$
$1 + x^{1 / 3} = t^{2}$
$x^{- 2 / 3} d x = 6 t d t$
$= \int 6 t^{2} d t$
$= 2 t^{3} + c$
$= 2 {(1 + x^{3})}^{3 / 2} + c$

Suggest Corrections

Similar questions

\int \frac{\sqrt{1 + \sqrt[3]{x}}}{\sqrt[3]{x^{2}}} d x

is equal to

Q. If

\int_{\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{x^{4}}{1 - x^{4}} {cos}^{- 1} \frac{2 x}{1 + x^{2}} d x = k \int_{0}^{\frac{1}{\sqrt{3}}} \frac{x^{4}}{1 - x^{4}} d x

then

^{'} k^{'}

is equal to?

\int \frac{1}{(x + 1) \sqrt{2 + x - x^{2}}} d x

is equal to

\int \frac{{sin}^{- 1} x}{\sqrt{1 - x^{2}}} d x

is equal to

\int \frac{{(x + \sqrt{1 + x^{2}})}^{15}}{\sqrt{1 + x^{2}}} d x

is equal to

Join BYJU'S Learning Program

∫√1+3√x3√x2dx is equal to

The correct option is C 2(1+x1/3)3/2+C∫√1+3√x3√x2dx=∫x−23(1+x13)1/2dx =∫x−2/3(1+x1/3)1/2dx 1+x1/3=t2 x−2/3dx=6 t dt =∫6 t2dt =2t3+c =2(1+x3)3/2+c

$\int \frac{\sqrt{1 + \sqrt[3]{x}}}{\sqrt[3]{x^{2}}} d x$ is equal to