Question
Evaluvate the following intregals
$\int\frac{2\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+5}}\text{dx}$

Answer

Let $\text{I}=\int\frac{2\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+5}}\text{dx}$
Let $2\text{x}+3=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+4\text{x}+5)+\mu$
$=\lambda(2\text{x}+4)+\mu$
$2\text{x}+3=(2\lambda)\text{x}+4\lambda+\mu$
Compairing the coefficient of like powers of x,
$2\lambda=2\ \Rightarrow\lambda=1$
$4\lambda+\mu=3\Rightarrow4(1)+\mu=3$
$\Rightarrow\mu=-1$
So, $\text{I}=\int\frac{(2\text{x}+4)-1}{\sqrt{\text{x}^2+4\text{x}+5}}\text{dx}$
$=\int\frac{(2\text{x}+4)}{\sqrt{\text{x}^2+4\text{x}+5}}\text{dx}-\int\frac{1}{\sqrt{\text{x}^2+2\text{x}(2)+(2)^2-(2)^2+5}}$
$=\int\frac{(2\text{x}+4)}{\sqrt{\text{x}^2+4\text{x}+5}}\text{dx}-\int\frac{1}{\sqrt{(\text{x}+2)^2+(1)^2}}\text{dx}$
$\text{I}=2\sqrt{\text{x}^2+4\text{x}+5}-\log\big|\text{x}+2+\sqrt{(\text{x}+2)^2+1}\big|+\text{C}$ 
$\Big[\text{since}, \int\frac{1}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\text{C},\int\frac{1}{\sqrt{\text{x}^2-\text{a}^2}}\text{dx}=\log\big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\big|+\text{C}\Big]$
$\text{I}=2\sqrt{\text{x}^2+4\text{x}+5}-3\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+5}\big|+\text{C}$

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