Mathematics

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

$\dfrac { 1 } { 3 } \ln \left| \frac { \sqrt { 1 - x ^ { 3 } } - 1 } { \sqrt { 1 - x ^ { 3 } } + 1 } \right| + C$

SOLUTION
$I=\displaystyle\int{\dfrac{dx}{x\sqrt{1-{x}^{3}}}}$

Let $u={x}^{3}\Rightarrow\,du=3{x}^{2}dx$ or $dx=\dfrac{du}{3{x}^{2}}$

$\Rightarrow\,I=\displaystyle\int{\dfrac{\dfrac{du}{3{x}^{2}}}{x\sqrt{1-{x}^{3}}}}$

$\Rightarrow\,I=\displaystyle\int{\dfrac{du}{3{x}^{3}\sqrt{1-{x}^{3}}}}$

$\Rightarrow\,I=\dfrac{1}{3}\displaystyle\int{\dfrac{du}{u\sqrt{1-u}}}$ where $u={x}^{3}$

Let $v=\sqrt{1-u}\Rightarrow\,dv=\dfrac{-du}{2\sqrt{1-u}}$ or $du=-2\sqrt{1-u}dv$

Again,$v=\sqrt{1-u}$

squaring both sides,${v}^{2}=1-u\Rightarrow\,u=1-{v}^{2}$

$\Rightarrow\,I=\dfrac{1}{3}\displaystyle\int{\dfrac{2\sqrt{1-u}dv}{\left({v}^{2}-1\right)\sqrt{1-u}}}$

$\Rightarrow\,I=\dfrac{2}{3}\displaystyle\int{\dfrac{dv}{\left(v-1\right)\left(v+1\right)}}$

$\Rightarrow\,I=\dfrac{2}{3\times 2}\displaystyle\int{\left[\dfrac{1}{v-1}-\dfrac{1}{v+1}\right]dv}$

$\Rightarrow\,I=\dfrac{1}{3}\left[\ln{\left|v-1\right|}-\ln{\left|v+1\right|}\right]+c$

$\Rightarrow\,I=\dfrac{1}{3}\ln{\left|\dfrac{v-1}{v+1}\right|}+c$

$\Rightarrow\,I=\dfrac{1}{3}\ln{\left|\dfrac{\sqrt{1-u}-1}{\sqrt{1-u}+1}\right|}+c$where ${v}^{2}=1-u$ and $v=\sqrt{1-u}$

$\Rightarrow\,I=\dfrac{1}{3}\ln{\left|\dfrac{\sqrt{1-{x}^{3}}-1}{\sqrt{1-{x}^{3}}+1}\right|}+c$ where $u={x}^{3}$

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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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