Question
Evaluate the following intregals:
$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}+1)^2}\ \text{dx}$
$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}+1)^2}\ \text{dx}$
✓
Answer
Let $\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{(\text{x}-1)}+\frac{\text{C}}{\text{(x}+1)^2}$
⇒ x2 = A(x + 1)2 + B(x - 1) (x + 1) + C(x - 1)
= (A + B) x2 + (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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