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

Question

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

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Answer

Let $\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{(\text{x}-1)}+\frac{\text{C}}{\text{(x}+1)^2}$
$\Rightarrow x^2 = A(x + 1)^2+ B(x - 1) (x + 1) + C(x - 1)$
$= (A + B) x^2 + (2A + C) x + (A - B - C)$
Equating similar terms,
A + B = , 2A + C = 0, A - B - C = 0
Solving, we get, $\text{A}=\frac{1}{4},\text{B}=\frac{3}{4},\text{C}=-\frac{1}{2}$
Thus,
$\text{I}=\frac{1}{4}\int\frac{\text{dx}}{\text{x}-1}+\frac{3}{4}\int\frac{\text{dx}}{\text{x}+1}-\frac{1}{2}\int\frac{\text{dx}}{(\text{x}+1)}^2$
$=\frac{1}{4}\log|\text{x}-1|+\frac{3}{4}\log|\text{x}+1|+\frac{1}{2(\text{x}+1)}+\text{C}$
$\text{I}=\frac{1}{4}\log|\text{x}-1|+\frac{3}{4}\log|\text{x}+1|+\frac{1}{2(\text{x}+1)}\text{C}$

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