Mathematics

$\int \dfrac{\sqrt [2] {x}}{1+\sqrt [4] {x^{3}}}dx$ is equal to

SOLUTION
We have,
$I=\int \dfrac{\sqrt [2] {x}}{1+\sqrt [4] {x^{3}}}dx$

Let
$t=1+x^{\dfrac{3}{4}}$
$\dfrac{dt}{dx}=0+\dfrac{3}{4}x^{\dfrac{-1}{4}}$

$dt=\dfrac{3}{4x^{\dfrac{1}{4}}}dx$

$dx=\dfrac{4x^{\dfrac{1}{4}}}{3}dt$

Therefore,
$I=\dfrac{4}{3}\displaystyle \int \dfrac{t-1}{t}dt$
$I=\dfrac{4}{3}\displaystyle \int \left( 1-\dfrac{1}{t}\right)dt$
$I=\dfrac{4}{3}\displaystyle \left( t-\ln t\right)+C$

On putting the value of $t$, we get
$I=\dfrac{4}{3}\displaystyle \left( 1+x^{\frac{3}{4}}-\ln (1+x^{\frac{3}{4}})\right)+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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