Question
Evaluate the following intregals:
$\int\frac{18}{(\text{x}+2)(\text{x}^2+4)}\text{ dx}$
$\int\frac{18}{(\text{x}+2)(\text{x}^2+4)}\text{ dx}$
$\Rightarrow18=\text{A}(\text{x}^2+4)+(\text{Bx}+\text{C})(\text{x}+2)$
$18=(\text{A}+\text{B})\text{x}^2+(2\text{B}+\text{C})\text{x}+(4\text{A}+2\text{C})$
Equating similar terms, get,
$\text{A}+\text{B}=0,2\text{B}+\text{C}=0,4\text{A}+2\text{C}=18$
Solving, we get,
$\text{A}=\frac{9}{4},\text{B}=-\frac{9}{4},\text{C}=\frac{9}{2}$
Thus,
$\text{I}=\frac{9}{4}\int\frac{\text{dx}}{\text{x}+2}+\Big(-\frac{9}{4}\Big)\frac{\text{x}}{\text{x}^2+4}\ \text{dx}+\frac{9}{2}\int\frac{\text{dx}}{\text{x}^2+4}$
$\text{I}=\frac{9}{4}\log|\text{x}+2|-\frac{9}{8}\log|\text{x}^2+4|+\frac{9}{4}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$ $\Big[\because\int\frac{\text{dx}}{\text{x}^2+\text{a}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{9}\Big]$
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