Evaluate the following intregals: x+2 x^2+2x-1 dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}-1}}\text{dx}$

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Answer

Let $\text{I}=\int\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}-1}}\text{dx}$
consider,
$\text{x}+2=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2+2\text{x}-1)+\text{B}$
$\Rightarrow\text{x}+2=\text{A}(2\text{x}+2)+\text{B}$
$\Rightarrow\text{x}+2=(2\text{A})\text{x}+2\text{A}+\text{B}$
Equating coefficient of like terms.
$2\text{A}=1$
$\Rightarrow\text{A}=\frac{1}{2}$
And
$2\text{A}+\text{B}=2$
$\Rightarrow2\times\frac{1}{2}+\text{B}=2$
$\Rightarrow\text{B}=1$
Then,
$\text{I}=\int\frac{\big[\frac{1}{2}(2\text{x}+2)+1}{\sqrt{\text{x}^2+2\text{x}+1}}\text{dx}$
$=\frac{1}{2}\int\frac{(2\text{x}+2)\text{dx}}{\sqrt{\text{x}^2+2\text{x}+1}}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}-1}}$
Let $\text{x}^2+2\text{x}-1=\text{t}$
$\Rightarrow(2\text{x}+2)\text{dx}=\text{dt}$
$\therefore\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}+1-2}}$
$=\frac{1}{2}\int\text{t}^{-\frac{1}{2}}\text{dt}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}+1-2}}$
$=\frac{1}{2}\bigg[\frac{\text{t}^{\frac{1}{2}}+1}{-\frac{1}{2}+1}\bigg]+\int\frac{\text{dx}}{\sqrt{(\text {x}+1)^2-(\sqrt{2})^2}}$
$=\sqrt{\text{t}}+\log\Big|\text{x}+1+\sqrt{(\text{x}+1)^2-(\sqrt{2})^2}\Big|+\text{C}$
$=\sqrt{\text{x}^2+2\text{x}-1}+\log\big|\text{x}+1+\sqrt{\text{x}^2+2\text{x}-1}\big|+\text{C}$

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