Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate the following integrals:$\int\frac{2\text{x}-3}{\text{x}^2+6\text{x}+13}\text{ dx}$
✓
Answer
Let $\text{I}=\int\frac{2\text{x}-3}{\text{x}^2+6\text{x}+13}\text{ dx}$
Let $2\text{x}-3=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2+6\text{x}+13\big)+\mu$
$=\lambda(2\text{x}+6)+\mu$
$2\text{x}-3=(2\lambda)\text{x}+(6\lambda+\mu)$
Comparing the coefficients of like powers of x,
$2\lambda=2\Rightarrow\lambda=1$
$6\lambda+\mu=-3\Rightarrow6(1)+\mu=-3$
$\mu=-9$
So, $\text{I}=\int\frac{1(2\text{x}+6)-9}{\text{x}^2+6\text{x}+13}\text{ dx}$
$\text{I}=\int\frac{2\text{x}+6}{\text{x}^2+6\text{x}+13}\text{ dx}-9\int\frac{1}{\text{x}^2+2\text{x}(3)+(3)^2-(3)^2+13}\text{ dx}$
$\text{I}=\int\frac{2\text{x}+6}{\text{x}^2+6\text{x}+13}\text{ dx}-9\int\frac{1}{(\text{x}+3)^2+(2)^2}\text{ dx}$
$\text{I}=\log\big|\text{x}^2+6\text{x}+13\big|-9\times\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}+3}{2}\Big)+\text{C}$ $\Big[\text{Since }\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\log\big|\text{x}^2+6\text{x}+13\big|-\frac{9}{2}\tan^{-1}\Big(\frac{\text{x}+3}{2}\Big)+\text{C}$
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