Question
Evaluate the following intregals:
$\int\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}\ \text{dx}$
$\int\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}\ \text{dx}$
✓
Answer
we have
$\text{I}=\int\frac{\text{x}\text{dx}}{(\text{x}-1)(\text{x}+2)}$
Let $\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{(\text{x}-1)^2}+\frac{\text{C}}{\text{x}+2}$
$\Rightarrow\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}=\frac{\text{A}(\text{x}-1)(\text{x}+2)+\text{B}(\text{x}+2)+\text{C}(\text{x}-1)^2}{(\text{x}-1)^2(\text{x}+2)}$
$\Rightarrow\text{x}=\text{A}(\text{x}^2+2\text{x}-\text{x}-2)+\text{B}(\text{x}+2)+\text{C}(\text{x}^2-2\text{x}+1)$
$\Rightarrow\text{x}=\text{A}(\text{x}^2+\text{x}-2)+\text{B}(\text{x}+2)+\text{C}(\text{x}^2-2\text{x}+1)$
$\Rightarrow\text{x}=(\text{A}+\text{C})\text{x}^2+(\text{A}+\text{B}-2\text{C})\text{x}+(-2\text{A}+2\text{B}+\text{C})$
A + C = 0 ...(1)
A + B - 2C = 1 ...(2)
-2A + 2B + C = 0 ...(3)
Solving(1), (2) and (3), we get
$\text{A}=\frac{2}{9},\text{B}=\frac{1}{3}\text{ and }\text{C}=-\frac{2}{9}$
$\therefore\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}=\frac{2}{9(\text{x}-1)}+\frac{1}{3(\text{x}-1)^2}-\frac{2}{9(\text{x}+2)}$
$\Rightarrow\text{I}=\frac{2}{9}\int\frac{\text{dx}}{\text{x}-1}+\frac{1}{3}\int\frac{\text{dx}}{(\text{x}-1)^3}-\frac{2}{9}\int\frac{\text{dx}}{\text{x}+2}$
$=\frac{2}{9}\log|\text{x}-1|+\frac{1}{3}\times\Big(\frac{-1}{\text{x}-1}\Big)-\frac{2}{9}\log|\text{x}+2 |+\text{C}$
$=\frac{2}{9}\log\Big(\frac{\text{x}-1}{\text{x}+2}\Big)-\frac{1}{3(\text{x}-1)}+\text{C}$
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