Evaluate the following intregals: 5x^2+20x+6 x^2+2x^2+x dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{5\text{x}^2+20\text{x}+6}{\text{x}^2+2\text{x}^2+\text{x}}\ \text{dx}$

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Answer

Let $\text{I}=\int\frac{5\text{x}^2+20\text{x}+6}{\text{x}^2+2\text{x}^2+\text{x}}\ \text{dx}$$=\int\frac{5\text{x}^2+20\text{x}+6}{\text{x}(\text{x}+1)^2}\ \text{dx}$
Now,
Let $\frac{5\text{x}^2+20\text{x}+6}{\text{x}(\text{x}+1)^2}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}+1}+\frac{\text{C}}{(\text{x}+1)^2}$
$\Rightarrow5\text{x}^2+20\text{x}+6=\text{A}(\text{x}+1)^2+\text{Bx}(\text{x}+1)+\text{Cx}$
Equating similar terms, we get,
A + B = 5, 2A + B + C = 20, A = 6
Solving, we get, B = -1, C = 9
Thus,
$\text{I}=\int\frac{6\text{dx}}{\text{x}}-1\int\frac{\text{dx}}{\text{x}+1}+9\int\frac{\text{dx}}{(\text{x}+1)^2}$
$\therefore\text{I}=6\log|\text{x}|-\log|\text{x}+1|-\frac{9}{\text{x}+1}+\text{C}$

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