Evaluate the following intregals: x x^2+6x+10 dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{\text{x}}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}$

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Answer

Let $\text{I}\int\frac{\text{x}}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}$
Let $\text{x}=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+6\text{x}+10)+\mu$
$=\lambda(2\text{x}+6)+\mu$
$\text{x}=(2\lambda)\text{x}+6\lambda+\mu$
Comparing the co-efficient of like powers of x.
$2\lambda=1\Rightarrow\lambda=\frac{1}{2}$
$6\lambda+\mu=0\Rightarrow6\Big(\frac{1}{2}\Big)+\mu=0$
$\mu=-3$
So, $\text{I}_1=\int\frac{\frac{1}{2}(2\text{x}+6)=3}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}$
$\frac{1}{2}\int\frac{{2\text{x}+6}=3}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}-3\text{ I }\frac{1}{\sqrt{\text{x}^2+2\text{x}(3)+(3)^2+10}}$
$\text{I}_1=\frac{1}{2}\int\frac{2\text{x}+6}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}-3\int\frac{1}{\sqrt{(\text{x}+3)^2+(1)^2}}\text{dx}$
$\text{I}=\sqrt{\text{x}^2+6\text{x}+10}-3\log\big|\text{x}+3+\sqrt{\text{x}^2+6\text{x}+10}\big|+\text{c}$
$\text{I}_1=\frac{1}{2}(2\sqrt{\text{x}^2+6\text{x}+10})-3\log\big|\text{x}+3+\sqrt{(\text{x}+3)^2+1}\big|+\text{c}$ $\Big[\text{since},\int\frac{1}{\sqrt{\text{x}}}\text{dx}-2\sqrt{\text{x}}+\text{c},\int\frac{1}{\sqrt{\text{x}^2+\text{a}^2}}\text{dx}-\log\big|\text{x}+\sqrt{\text{x}}^2+\text{a}^2\big|+\text{c}\Big]$
$\text{I}=\sqrt{\text{x}^2+6\text{x}+10}-3\log\big|\text{x}+3+\sqrt{\text{x}^2+6\text{x}+10}\big|+\text{c}$

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