91Ó°ÊÓ

From the expression of the function, it can be seen that a suitable substitution would be: $$x = 6 \cosh{t}$$ This is because \(\cosh^2{t} - 1 = \sinh^2{t}\). And since x > 6, t > 0. Now, we find the derivative, \(dx = 6\sinh{t} \, dt\).

We now substitute the expressions for \(x\) and \(dx\) in the integral and simplify: $$\int \frac{6\sinh{t}}{\left(6^{2}(\cosh^2{t} - 1)\right)^{3 / 2}}dt$$

Now, we need to simplify the expression inside the integral: $$\int \frac{6\sinh{t}}{\left(36(\sinh^2{t})\right)^{3 / 2}} dt$$ $$\int \frac{6\sinh{t}}{216(\sinh^3{t})} dt$$ $$\int \frac{1}{36\sinh^2{t}} dt$$

The integral can now be solved: $$\int \frac{1}{36\sinh^2{t}} dt = \frac{1}{36} \int \cosh^2{t} dt$$ To solve the integral of \(\cosh^2{t}\), we use the double angle formula: $$\cosh(t)=\frac{1}{2}(1+\cos{(2t)})$$ Then the integral becomes $$\frac{1}{36} \int\left (\frac{1}{2}(1+\cos{(2t)})\right) dt$$ Integrate with respect to t: $$\frac{1}{36}\left(\frac{1}{2}t+\frac{1}{4}\sin{(2t)}\right)+C$$

Now, we need to substitute back the original variable, x. From our initial substitution: $$x=6\cosh{t}\Rightarrow t=\cosh^{-1}\left(\frac{x}{6}\right)$$ And $$2t=2\cosh^{-1}\left(\frac{x}{6}\right)=\cosh^{-1}\left(\frac{x^2}{36}\right)$$ $$\sin{(2t)}=\sin{(\cosh^{-1}(\frac{x^2}{36}))}=\frac{36-\frac{x^2}{36}}{x} = \frac{1296-x^2}{36x}$$ Finally, we plug these expressions back into the integral: $$\frac{1}{36}\left(\frac{1}{2}\cosh^{-1}\left(\frac{x}{6}\right)+\frac{1}{4}\frac{1296-x^2}{36x}\right)+C$$ And simplify: $$\frac{1}{72}\cosh^{-1}\left(\frac{x}{6}\right)+\frac{1}{144}\frac{1296-x^2}{x}+C$$ Thus, the evaluated integral is: $$\int \frac{d x}{\left(x^{2}-36\right)^{3 / 2}} = \frac{1}{72}\cosh^{-1}\left(\frac{x}{6}\right)+\frac{1}{144}\frac{1296-x^2}{x}+C$$