Mathematics

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


ANSWER

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


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,

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

Hence, this is the answer
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