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Step-by-step Solution

Integral of $\frac{1}{x^3\sqrt{x^2-16}}$ with respect to x

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Answer

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

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

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

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Answer

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

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