How to calculate the integral $\int\frac{1}{\sqrt{(x^2+8)^3}}dx$? I need to solve something like this
$$\int\frac{1}{\sqrt{(x^2+8)^3}}dx$$
Wolfram alpha says the solution is $$\frac{x}{8\sqrt{x^2+8}} + c$$
The problem is that the integrand is obtained by the quotient rule:
$$\bigg(\frac{g(x)}{h(x)}\bigg)'=\frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)}$$
$$\bigg(\frac{x}{8\sqrt{x^2+8}}\bigg)'=\frac{1}{8}\cdot\frac{\sqrt{x^2+8}-\frac{x^2}{\sqrt{x^2+8}}}{x^2+8}=\frac{1}{8}\cdot\frac{\frac{x^2+8-x^2}{\sqrt{x^2+8}}}{x^2+8}=\frac{1}{8}\cdot\frac{8}{\sqrt{(x^2+8)^3}}=\frac{1}{\sqrt{(x^2+8)^3}}$$
It there a way to extract the solution from these types of integrals which argument is born from the easy quotient rule?

As mentioned in another answer, the solution to the given integral can be found by making a substitution with a trigonometric function, then integrating, calculating, substituting back and simplifying.
For a general integral of the form
$I =\displaystyle \int \frac{1}{\left(a x^n+b\right)^p} \, dx$ ,
the solution can be obtained with the help of Mathematica in terms of a special function:
$\displaystyle I = x \left(a x^n+b\right)^{-p} \left(\frac{a x^n}{b}+1\right)^p \, _2F_1\left(\frac{1}{n},p;1+\frac{1}{n};-\frac{a x^n}{b}\right)+C$,
where $\, _2F_1(a,b;c;z)$ is the Gaussian hypergeometric function.
The integral in this question is the special case where
$a =1, b = 8, n = 2,p =3/2$
A function can be defined with Mathematica:
solh[a_, b_, n_, 
   p_] = (x*(1 + (a*x^n)/b)^p*
     Hypergeometric2F1[1/n, p, 1 + 1/n, -((a*x^n)/b)])/
       (b + a*x^n)^p; 

Then taking the special values for the integral in this question:
FullSimplify[solh[1, 8, 2, 3/2]]

It can be verified that the solution is
$\displaystyle \frac{x}{8\sqrt{x^2+8}} + constant$
