Integrate the function in Exercise: 1 x-x^ 3 - Vidyadip

Question

Integrate the function in Exercise:
$\frac{1}{\text{x}-\text{x}^{3}}$

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Answer

$\frac{1}{\text{x}-\text{x}^{3}}=\frac{1}{\text{x}(1-\text{x}^{2})}=\frac{1}{\text{x}(1-\text{x})(1+\text{x})}$ $\text{Let}\frac{1}{\text{x}(1-\text{x})(1+\text{x)}}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{(1-\text{x})}+\frac{\text{C}}{1+\text{x}}$ $\Rightarrow1=\text{A}(1-\text{x}^{2})+\text{B}\text{x}(1+\text{x})+\text{c}\text{x}(1-\text{x})$$\Rightarrow1=\text{A}-\text{A}\text{x}^{2}+\text{B}\text{x}+\text{B}\text{x}^{2}+\text{C}\text{x}-\text{C}\text{x}^{2}$
Equating the coefflclents of $\text{x}^{2},\text{x}$ and constant term, we obtain $-\text{A}+\text{B}-\text{C}=0$ $\text{B}+\text{C}=0$ $\text{A}=1$ on solving these equations, we obtain$\text{A}=1,\text{B}=\frac{1}{2},\text{and}\ \text{C}=\frac{1}{2}$
from eqution (1), we obtain $\frac{1}{\text{x}(1-\text{x})(1+\text{x})}=\frac{1}{\text{x}}+\frac{1}{2(1-\text{x})}-\frac{1}{2(1+\text{x})}$ $\Rightarrow\int\frac{1}{\text{x}(1-\text{x})(1+\text{x}}\text{dx}=\int\frac{1}{\text{x}}\text{dx}+\frac{1}{2}\int\frac{1}{1-\text{x}}\text{dx}-\frac{1}{2}\int\frac{1}{1+\text{x}}\text{dx}$ $=\log|\text{x}|-\frac{1}{2}\log|(1-\text{x})|-\frac{1}{2}\log|(1+\text{x)}|$ $=\log|\text{x}|-\log\bigg|(1-\text{x})^{\frac{1}{2}}\bigg|-\log|(1+\text{x)}^{\frac{1}{2}}\bigg|$ $=\log\left|\frac{\text{x}}{(1-\text{x)}^{\frac{1}{2}}(1+\text{x)}^{\frac{1}{2}}}\right|+\text{C}$ $=\log\left|\bigg(\frac{\text{x}^{2}}{1-\text{x}^{2}}\bigg)^{\frac{1}{2}}\right|+\text{C}$ $=\frac{1}{2}\log\left|\frac{\text{x}^{2}}{1-\text{x}}\right|+\text{C}$

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