Evaluate the following integrals: 1 (x^2-1) x^2+1 dx - Vidyadip

Question

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

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Answer

Let $\text{I}=\int\frac{1}{(\text{x}^2-1)\sqrt{\text{x}^2+1}}\text{ dx}$
Let $\text{x}=\frac{1}{\text{t}}$
$\text{dx}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=-\int\frac{\frac{1}{\text{t}^2}\text{ dt}}{\Big(\frac{1}{\text{t}^2}-1\Big)\sqrt{\Big(\frac{1}{\text{t}^2}+1\Big)}}$
$=-\int\frac{\text{t dt}}{(1-\text{t}^2)\sqrt{1+\text{t}^2}}$
Let $1+\text{t}^2=\text{u}^2$
$2\text{tdt}=2\text{udt}$
$\text{I}=\int\frac{\text{udu}}{(\text{u}^2-2)\text{u}}$
$=\int\frac{\text{du}}{\text{u}^2-2}$
$\therefore\ \text{I}=\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{u}-\sqrt{2}}{\text{u}+\sqrt{2}}\bigg|+\text{C}$
$=\frac{1}{2\sqrt{2}}\log\bigg|\frac{\sqrt{1+\text{t}^2}-\sqrt{2}}{\sqrt{1+\text{t}^2}+\sqrt{2}}\bigg|+\text{C}$
Hence,
$\text{I}=-\frac{1}{2\sqrt{2}}\log\bigg|\frac{\sqrt{2}\text{x}+\sqrt{\text{x}^2+1}}{\sqrt{2}\text{x}-\sqrt{\text{x}^2+1}}\bigg|+\text{C}$

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