Evaluvate the following intregals 2x+1 x^2+4x+3 dx

Question

Evaluvate the following intregals
$\int\frac{2\text{x}+1}{\sqrt{\text{x}^2+4\text{x}+3}}\text{dx}$

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Answer

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

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