Evaluate the follwing intregals: 1 x^4-1 dx - Vidyadip

Question

Evaluate the follwing intregals:
$\int\frac{1}{\text{x}^4-1}\text{ dx}$

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Answer

The Evaluate the integral follow the steps,
$\int\frac{1}{(\text{x}^4-1)}\text{ dx}$
Let $\frac{1}{(\text{x}^4-1)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}-1}+\frac{\text{C}}{\text{x}^2+1}$
$1={\text{A}}{(\text{x}-1})(\text{x}^2+1)+\text{B}(\text{x}+1)(\text{x}^2+1)+\text{C}(\text{x}+1)(\text{x}-1)$
$\text{For x}=1,\text{ B}=\frac{1}{4}$
$\text{For x}=-1,\text{ A}=-\frac{1}{4}$
$\text{For x}=0,\text{ C}=-\frac{1}{2}$
Therefore,
$\int\frac{1}{(\text{x}^4-1)}\text{ dx}=-\frac{1}{4}\int\frac{\text{dx}}{\text{x}+1}+\frac{1}{4}\frac{\text{dx}}{\text{x}-1}-\frac{1}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
$=-\frac{1}{4}\ln\big|(\text{x}+1)\big|+\frac{1}{4}\ln\big|(\text{x}-1)\big|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
$=\frac{1}{4}\ln\Big|\frac{\text{x}-1}{\text{x}+1}\Big|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$

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