Mathematics

# The value of $\int \dfrac{dx}{x\sqrt{1-x^{3}}}$ is equal

$I=\dfrac{-2}{3}{{\tan }^{-1}}\sqrt{1-{{x}^{3}}}+C$

##### SOLUTION

We have,

$I=\int{\dfrac{dx}{x\sqrt{1-{{x}^{3}}}}}\,\,......\,\left( 1 \right)$

Let,

$u=\sqrt{1-{{x}^{3}}}$

$\dfrac{du}{dx}=\dfrac{1}{2\sqrt{1-{{x}^{3}}}}\left( -3{{x}^{2}} \right)$

$\dfrac{du}{dx}=-\dfrac{3}{2u}{{x}^{2}}$

$\dfrac{dx}{du}=-\dfrac{2u}{3{{x}^{2}}}$

$dx=-\dfrac{2u}{3{{x}^{2}}}du$

Then,

Put the value of $dx$ in equation (1) and we get,

$I=\int{\dfrac{1}{xu}\times -\dfrac{2u}{3{{x}^{2}}}du}$

$I=\int{-\dfrac{2}{3{{x}^{3}}}du}$

$I=\dfrac{-2}{3}\int{\dfrac{du}{{{x}^{3}}}}$

$I=\dfrac{-2}{3}\int{\dfrac{du}{{{u}^{2}}+1}}$

$I=\dfrac{-2}{3}\int{\dfrac{du}{{{u}^{2}}+{{1}^{2}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore I=\int{\dfrac{dx}{1+{{x}^{2}}}}={{\tan }^{-1}}x$

$I=\dfrac{-2}{3}{{\tan }^{-1}}u+C$

Put the value of u and we get,

$I=\dfrac{-2}{3}{{\tan }^{-1}}\sqrt{1-{{x}^{3}}}+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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