# Step-by-step Solution

## Integral of (x^2)/((x^2-4)^0.5)

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$4\left(\frac{\frac{1}{2}\left(\frac{x}{2}\right)^{2}\sqrt{x^2-4}}{x}+\frac{1}{2}\ln\left|\frac{x+\sqrt{x^2-4}}{2}\right|\right)+C_0$

## Step-by-step explanation

Problem to solve:

$\int\frac{x2}{\sqrt{x^2-4}}dx$
1

Solve the integral $\int\frac{x^2}{\sqrt{x^2-4}}dx$ by trigonometric substitution using the substitution

$\begin{matrix}x=2\sec\left(\theta\right) \\ dx=2\sec\left(\theta\right)\tan\left(\theta\right)d\theta\end{matrix}$
2

Substituting in the original integral, we get

$\int\frac{8\sec\left(\theta\right)^2\tan\left(\theta\right)\sec\left(\theta\right)}{\sqrt{4\sec\left(\theta\right)^2-4}}d\theta$

$4\left(\frac{\frac{1}{2}\left(\frac{x}{2}\right)^{2}\sqrt{x^2-4}}{x}+\frac{1}{2}\ln\left|\frac{x+\sqrt{x^2-4}}{2}\right|\right)+C_0$
$\int\frac{x2}{\sqrt{x^2-4}}dx$