Evaluate the follwing intregals: x^2 (x-1)(x^2+1) dx - Vidyadip

Question

Evaluate the follwing intregals:
$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}^2+1)}\ \text{dx}$

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Answer

$\text{I}=\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}^2+1)}\ \text{dx}$
$\frac{\text{x}^2}{(\text{x}-1)^2(\text{x}^2+1)}=\frac{\text{A}}{(\text{x}-1)}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$=\frac{\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}-1)}{(\text{x}-1)(\text{x}^2+1)}$
$\Rightarrow\frac{\text{x}^2}{(\text{x}+1)(\text{x}^2+1)}=\frac{(\text{A}+\text{B})\text{x}^2+(\text{C}-\text{B})\text{x}+(\text{A}-\text{C})}{(\text{x}-1)(\text{x}^2+1)}$
Comapairing coefficient, we get
$\text{A}+\text{B}=\text{C}=\frac{1}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{1}{(\text{x}-1)\ \text{dx}}+\frac{1}{2}\int\frac{\text{x}}{\text{x}^2+1}\ \text{dx}+\frac{1}{2}\int\frac{1}{\text{x}^2+1}\ \text{dx}$
$=\frac{1}{2}\ln|\text{x}-1|+\frac{1}{4}\ln|\text{x}^2+1|+\frac{1}{2}\tan^{-1}\text{(x)}+\text{C}$

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