# Step-by-step Solution

## Integral of 1/(x^3(x^2-16)^0.5)

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$\frac{1}{128}arcsec\left(\frac{x}{4}\right)+\frac{1}{32}\left(\frac{\sqrt{x^2-16}}{x^2}\right)+C_0$

## Step-by-step explanation

Problem to solve:

$\int\frac{dx}{x^3\sqrt{x^2-16}}$
1

Solve the integral $\int\frac{1}{x^3\sqrt{x^2-16}}dx$ by trigonometric substitution using the substitution

$\begin{matrix}x=4\sec\left(\theta\right) \\ dx=4\sec\left(\theta\right)\tan\left(\theta\right)d\theta\end{matrix}$
2

Substituting in the original integral, we get

$\int\frac{4\tan\left(\theta\right)\sec\left(\theta\right)}{64\sec\left(\theta\right)^3\sqrt{16\sec\left(\theta\right)^2-16}}d\theta$

$\frac{1}{128}arcsec\left(\frac{x}{4}\right)+\frac{1}{32}\left(\frac{\sqrt{x^2-16}}{x^2}\right)+C_0$

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$\int\frac{dx}{x^3\sqrt{x^2-16}}$

### Main topic:

Integration by trigonometric substitution

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