Question
Evaluate the following intregals:
$\int\frac{5\text{x}^2-1}{\text{x}(\text{x}-1)(\text{x}+1)}\ \text{dx}$
$\int\frac{5\text{x}^2-1}{\text{x}(\text{x}-1)(\text{x}+1)}\ \text{dx}$
✓
Answer
Let $\int\frac{5\text{x}^2-1}{\text{x}(\text{x}-1)(\text{x}+1)}\ \text{dx}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}-1}+\frac{\text{C}}{\text{x}+1}$
$\Rightarrow5\text{x}^2-1=\text{A}(\text{x}^2-1)+\text{B}(\text{x}+1)\text{x}+\text{C}(\text{x}-1)\text{x}$
Put x = 0
⇒ -1 = -A ⇒ A = 1
Put x = +1
⇒ 4 = 2B ⇒ B = 2
Put x = -1
⇒ 4 = 2C ⇒ C = 2
So,
$\text{I}=\int\frac{\text{dx}}{\text{x}}+\int\frac{2\text{dx}}{\text{x}-1}+\int\frac{2\text{dx}}{\text{x}+1}$
$=\log|\text{x}|+2\log|\text{x}-1|+2\log|\text{x}+1|+\text{C}$
$\text{I}=\log|\text{x}(\text{x}^2-1)^2|$
$\Rightarrow5\text{x}^2-1=\text{A}(\text{x}^2-1)+\text{B}(\text{x}+1)\text{x}+\text{C}(\text{x}-1)\text{x}$
Put x = 0
⇒ -1 = -A ⇒ A = 1
Put x = +1
⇒ 4 = 2B ⇒ B = 2
Put x = -1
⇒ 4 = 2C ⇒ C = 2
So,
$\text{I}=\int\frac{\text{dx}}{\text{x}}+\int\frac{2\text{dx}}{\text{x}-1}+\int\frac{2\text{dx}}{\text{x}+1}$
$=\log|\text{x}|+2\log|\text{x}-1|+2\log|\text{x}+1|+\text{C}$
$\text{I}=\log|\text{x}(\text{x}^2-1)^2|$
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