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

Question

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

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Answer

We have$\text{I}=\int\frac{(\text{x}^2+\text{x}+1)}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}$
Let, $\int\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{C}}{\text{x}+2}$
$\Rightarrow\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}=\frac{\text{A}(\text{x}+1)(\text{x}+2)+\text{B}(\text{x}+2)+\text{C}(\text{x}+1)^2}{(\text{x}+1)^2(\text{x}+2)}$
$\Rightarrow\text{x}^2+\text{x}+1=\text{A}(\text{x}^2+\text{x}+2\text{x}+2)\\+\text{Bx}+2\text{B}+\text{C}(\text{x}^2+2\text{x}+1)$
$\Rightarrow\text{x}^2+\text{x}+1=(\text{A}+\text{C})\text{x}^2+(3\text{A}+\text{B}+2\text{C})\text{x}\\+(2\text{A}+2\text{B}+\text{C})$
Equating coefficient of like terms,
$\text{A}+\text{C}=1\ ...(1)$
$3\text{A}+\text{B}+2\text{C}=1\ ...(2)$
$2\text{A}+2\text{B}+\text{C}=1\ ...(3)$
Solving these three equation we get
$\text{A}=-2$
$\text{B}=1$
$\text{C}=3$
Hence, $\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}=\frac{-2}{\text{x}+1}+\frac{1}{(\text{x}+1)^2}+\frac{3}{\text{x}+2}$
$\therefore\text{I}=-2\int\frac{\text{dx}}{\text{x}+1}+\int\frac{\text{d}}{(\text{x}+1)^2}+3\int\frac{\text{dx}}{\text{x}+2}$
$=-2\log|\text{x}+1|-\frac{1}{\text{x}+1}+3\log|\text{x}+2|+\text{C}$

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