Question
Evaluate the following intregals:
$\int\frac{\text{x}^2+\text{x}-1}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}$
$\int\frac{\text{x}^2+\text{x}-1}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}$
✓
Answer
Let $\int\frac{\text{x}^2+\text{x}-1}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}=\frac{\text{A}}{(\text{x}+1)}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{C}}{(\text{x}+2)}$
$\Rightarrow\text{x}^2+\text{x}-1=\text{A}(\text{x}+1)(\text{x}+2)+\text{B}(\text{x}+2)+\text{C}(\text{x}+1)^2$
$=(\text{A}+\text{C})\text{x}^2+(3\text{A}+\text{B}+2\text{C})\text{x}+(2\text{A}+2\text{B}+\text{C})$
Equating similar terms
$\text{A}+\text{C}=1,3\text{A}+\text{B}+2\text{C}=1,2\text{A}+2\text{B}+\text{C}=-1$
Solving, we get, A = 0, B = -1, C = 1
Thus,
$\text{I}=0\int\frac{\text{dx}}{\text{x}+1}+(-1)\int\frac{\text{dx}}{(\text{x}+1)^2}+1\int\frac{\text{dx}}{(\text{x}+2)}$
$=\frac{1}{\text{x}+1}+\log|\text{x}+2|+\text{C}$
$\text{I}=\frac{1}{\text{x}+1}+\log|\text{x}+2|+\text{C}$
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