How to calculate the integral $\int\frac{1}{\sqrt{(x^2+8)^3}}dx$?

Question

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?

Answer 1

substitute $$x=2\sqrt{2}\tan(u)$$ then we get $$dx=2\sqrt{2}\sec^2(u)du$$

Answer 2

It there a way to extract the solution from these types of integrals which argument is born from the easy quotient rule?

None that I know of. You just see it or maybe make a lucky substitution.

Alternatively, you can take a look at Euler substitutions, which will work in this case and can help for quite a lot of hand-made integrals for homeworks.

Answer 3

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:

1. If $n=1$, put $u=ax+b$. ($n=0$ is trivial.)
2. If $n=2$, put $x=\left(\frac{b}{a}\right)^\frac{1}{n} tan u$.
3. 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.

Answer 4

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

$\displaystyle \frac{x}{8\sqrt{x^2+8}} + constant$