Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS3 Marks
Question
Integrate the function in Exercise:$\frac{\text{x}^{2}+\text{x}+1}{(\text{x}+1)^{2}(\text{x}+2)}$
✓
Answer
$\text{Let}\frac{\text{x}^{2}+\text{x}+1}{(\text{x}+1)^{2}(\text{x}+2)}=\frac{\text{A}}{\text{(x}+1)}+\frac{\text{B}}{\text{(x}+1)^{2}}+\frac{\text{C}}{\text{(x}+2)}$
$\Rightarrow{\text{x}^{2}+\text{x}+1}=\text{A}(\text{x}+1)\text{(x}+2)+\text{B}(\text{x}+2)+\text{C}(\text{x}^{2}+2\text{x}+1)$
$\Rightarrow{\text{x}^{2}+\text{x}+1}=\text{A}(\text{x}^{2}+3\text{x}+2)+\text{B}(\text{x}+2)+\text{C}(\text{x}^{2}+2\text{x}+1)$
$\Rightarrow{\text{x}^{2}+\text{x}+1}=\text{(A}+\text{C)}\text{x}^{2}+(3\text{A}+\text{B}+2\text{C})\text{x}+(2\text{A}+2\text{B}+\text{C)}$
Equating the cofficients of $\text{x}^{2},\text{x,}$ and constant term, we obtain
$\text{A}+\text{C}=1$
$3\text{A}+\text{B}+2\text{C}=1$
$2\text{A}+2\text{B}+\text{C}=1$
On solving these equations, we obtain
$\text{A}=-2,\text{B}=1,$ and $\text{C}=3$
From equation (1), we obtain
$\frac{\text{x}^{2}+\text{x}+1}{\text{(x}+1)^{2}\text{(x}+2)}=\frac{-2}{\text{(x}+1)}+\frac{3}{\text{(x}+2)}+\frac{1}{\text{(x}+1)^{2}}$
$\int\frac{\text{x}^{2}+\text{x}+1}{\text{(x}+1)^{2}\text{(x}+2)}\text{dx}=-2\int\frac{1}{\text{x}+1}\text{dx}+3\int\frac{1}{\text{(x}+2)}\text{dx}+\int\frac{1}{\text{(x}+1)^{2}}\text{dx}$
$=-2\log|\text{x}+1|+3\log|\text{x}+2|-\frac{1}{\text{(x}+1)^{2}}\text{dx}$
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