<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] I don't know of any general formula for the quotient rule.
But for an integral of the form
$$\int \frac{dx}{\left(ax^n+b\right)^{k}}$$
You can try the following substitutions:


*

*If $n=1$, put $u=ax+b$. ($n=0$ is trivial.)

*If $n=2$, put $x=\left(\frac{b}{a}\right)^\frac{1}{n} tan u$.

*If $n\ge 3$, put $x=\left(\frac{b}{a}\right)^\frac{1}{n} u$ and then factorise the polynomial in the denominator into linear and quadratic terms and then, after using the method of partial fractions to separate the terms, apply 1 and 2 as above.


This is just a general procedure, and may not be applicable always. Some problems will possess solutions obtained only by special methods.
Hope this helps.
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