# Thread: Integration of 1/(1+x^4)^(1/4)

1. ## Integration of 1/(1+x^4)^(1/4)

So I have been having trouble with finding the indefinite integral of
1/ (1+x^4)^(1/4), by this I mean (1+x^4)^(-1/4).
I was given a hint that says to use the substitution t^4= 1+x^(-1/4).
Any help would be appreciated, and thanks in advance!

2. Wolfram says that the solution involves the Hypergeometric Function.

http://www.wolframalpha.com/input/?i=integral[%281+%2B+x^4%29^%28-1%2F4%29]

3. I tell you

There exists a solution , perhaps the only one , to this integral

$\int \frac{dx}{ (1 + x^n)^{\frac{1}{n}}}$

Which is substituting $1 + x^{-n} = t^n$ <- be cafeful

the power of $x$ is this time $-n$ not $n$

so

$-n x^{-n-1}dx = n t^{n-1} dt$

$dx = - x^{n+1} t^{n-1} dt$

$\int \frac{dx}{ (1 + x^n)^{\frac{1}{n}}}$

$\int \frac{1}{ (1 + x^n)^{\frac{1}{n}}} \cdot ( - x^{n+1} t^{n-1} dt )$

$= - \int \frac{x^{n+1} t^{n-1}}{x (1 + x^{-n})^{\frac{1}{n}}} ~dt$

$- \int \frac{x^{n} t^{n-1}}{ t} ~dt$

$- \int \frac{t^{n-2}}{t^n - 1} ~dt$

When $n = 4$

the integral is tansformed to

$- \int \frac{ t^2 }{ t^4 - 1}~dt$

$= -\frac{1}{2} \int \left [ \frac{1}{t^2 - 1} + \frac{1}{t^2 + 1} \right ]~dt$

4. Originally Posted by Prove It
Wolfram says that the solution involves the Hypergeometric Function.

http://www.wolframalpha.com/input/?i=integral[%281+%2B+x^4%29^%28-1%2F4%29]
You don't always have to believe to Wolfram.

In the integral put $x=\frac1z$ and it becomes $-\int\frac{dz}{z\sqrt[4]{1+z^4}},$ now put $t^4=1+z^4$ and the integral is $-\int\frac{t^2}{t^4-1}\,dt$ which is the same as simplependulum gave.