Question
Evaluate the following integrals:
$\int\frac{\text{x}^2}{\text{x}^2+6\text{x}+12}\text{ dx}$
✓
Answer
Let $\text{I}=\int\frac{\text{x}^2\text{ dx}}{\text{x}^2+6\text{x}+12}$ Now,
Therefore,
$\frac{\text{x}^2}{\text{x}^2+6\text{x}+12}=1-\frac{(6\text{x}+12)}{\text{x}^2+6\text{x}+12}\ ....(1)$Let
$6\text{x}+12=\text{A}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+6\text{x}+12\big)+\text{B}$$\Rightarrow6\text{x}+12=\text{A}(2\text{x}+6)+\text{B}$
$\Rightarrow6\text{x}+12=(2\text{A})\text{x}+6\text{A}+\text{B}$
Equating coefficients of like terms $2\text{A}=6$ $\text{A}=3$ $6\text{A}+\text{B}=12$ $18+\text{B}=12$ $\text{B}=-6$ $\therefore\ \frac{\text{x}^2}{\text{x}^2+6\text{x}+12}=1-\frac{3(2\text{x}+6)}{\text{x}^2+6\text{x}+12}$ $\text{I}=\int\frac{\text{x}^2\text{ dx}}{\text{x}^2+6\text{x}+12}$ $=\int\text{dx}-3\int\frac{(2\text{x}+6)\text{ dx}}{\text{x}^2+6\text{x}+12}+6\int\frac{\text{dx}}{\text{x}^2+6\text{x}+12}$ $=\int\text{dx}-3\int\frac{(2\text{x}+6)\text{ dx}}{\text{x}^2+6\text{x}+12}+6\int\frac{\text{dx}}{\text{x}^2+6\text{x}+9+3}$ $=\int\text{dx}-3\int\frac{(2\text{x}+6)\text{ dx}}{\text{x}^2+6\text{x}+12}+6\int\frac{\text{dx}}{(\text{x}+3)^2+\big(\sqrt3\big)^2}$ $=\text{x}-3\log\big|\text{x}^2+6\text{x}+12\big|+\frac{6}{\sqrt3}\tan^{-1}\Big(\frac{\text{x}+3}{\sqrt3}\Big)+\text{C}$ $=\text{x}-3\log\big|\text{x}^2+6\text{x}+12\big|+2\sqrt3\tan^{-1}\Big(\frac{\text{x}+3}{\sqrt3}\Big)+\text{C}$Need a full question paper?
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