<s>[INST] 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?
 [/INST] 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
$\displaystyle \frac{x}{8\sqrt{x^2+8}} + constant$
</s>