Mathematics

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


ANSWER

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


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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