Evaluate the following integrals: 1 (2x^2+3) x^2-4 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1}{(2\text{x}^2+3)\sqrt{\text{x}^2-4}}\text{ dx}$

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Answer

Let $\text{I}=\int\frac{1}{(2\text{x}^2+3)\sqrt{\text{x}^2-4}}\text{ dx}$
Let $\text{x}=\frac{1}{\text{t}}$
$\text{dx}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=\int\frac{-\frac{1}{\text{t}^2}\text{ dt}}{\Big(\frac{2}{\text{t}^2+3}\Big)\sqrt{\big(\frac{1}{\text{t}^2}-4\big)}}$
$=-\int\frac{\text{t dt}}{(2+3\text{t}^2)\sqrt{1-4\text{t}^2}}$
Let $1-4\text{t}^2=\text{u}^2$
$-8\text{tdt}=2\text{udu}$
$\therefore\ \text{I}=\frac{1}{4}\int\frac{\text{u du}}{\frac{(11-3\text{u})^2}{4}\text{u}}$
$=\frac{1}{3}\int\frac{\text{du}}{\frac{11}{3}-\text{u}^2}$
$=\frac{1}{2\sqrt{33}}\log\begin{vmatrix}\frac{\text{u}-\sqrt{\frac{11}{3}}}{\text{u}+\sqrt{\frac{11}{3}}}+\text{C}\end{vmatrix}$
$=\frac{1}{2\sqrt{33}}\log\begin{vmatrix}\frac{\sqrt{1-4\text{t}^2}-\sqrt{\frac{11}{3}}}{\sqrt{1-4\text{t}^2}+\frac{11}{3}}+\text{C}\end{vmatrix}$
Hence,
$\text{I}=\frac{1}{2\sqrt{33}}\log\bigg|\frac{\sqrt{11}\text{x}+\sqrt{3\text{x}^2-12}}{\sqrt{11}\text{x}-\sqrt{3\text{x}^2-12}}\bigg|+\text{C}$

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