Step-by-step Solution

Integral of $\frac{1}{\sqrt{x^2-36}}$ with respect to x

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$\ln\left|\frac{x+\sqrt{x^2-36}}{6}\right|+C_0$

Step-by-step explanation

Problem to solve:

$\int\frac{1}{\sqrt{x^2-36}}dx$
1

Solve the integral $\int\frac{1}{\sqrt{x^2-36}}dx$ by trigonometric substitution using the substitution

$\begin{matrix}x=6\sec\left(\theta\right) \\ dx=6\sec\left(\theta\right)\tan\left(\theta\right)d\theta\end{matrix}$
2

Substituting in the original integral, we get

$\int\frac{6\tan\left(\theta\right)\sec\left(\theta\right)}{\sqrt{36\sec\left(\theta\right)^2-36}}d\theta$

$\ln\left|\frac{x+\sqrt{x^2-36}}{6}\right|+C_0$
$\int\frac{1}{\sqrt{x^2-36}}dx$