Integration change of variable I came across this problem and I have no idea how to solve it. I'd really appreciate your help if you could shed some light on it.

$$\int_0^\infty\left(\frac {x^3}{e^x-1}\right)\;dx$$
Perform a change of variable:
$$z = \frac{x}{1+x}$$
so that the integral reads
$$\int_0^1f(z)\;dz$$


$$ \int_0^{\infty} \frac{x^3}{e^x-1}dx$$
Let $z=\frac{x}{x+1}$, so now the lower and upper bounds are 
$$ \lim\limits_{x\to 0} \frac{x}{x+1} = \frac{0}{0+1} = \frac{0}{1} = 0 $$
And
$$ \lim\limits_{x\to\infty} \frac{x}{x+1} = \lim\limits_{x\to\infty} \frac{1}{1+\frac{1}{x}} = \frac{1}{1+0}=1$$
Also note that
$$z=\frac{x}{x+1} $$
$$z(x+1)=x $$
$$zx+z=x $$
$$z=x-zx $$
$$z=x(1-z) $$
$$x=\frac{z}{1-z} $$
And
$$ dz = \frac{1}{(x+1)^2}dx $$
$$ (x+1)^2 dz = dx $$
$$ \left(\frac{z}{1-z}+1\right)^2 dz = dx $$
$$ \left(\frac{z+1-z}{1-z}\right)^2 dz = dx $$
$$ \left(\frac{1}{1-z}\right)^2 dz = dx $$
$$ \frac{1}{(1-z)^2} dz = dx $$
So now we have
$$ \int_0^1 \frac{\left(\frac{z}{1-z}\right)^3}{\left(\exp\left(\frac{z}{1-z}\right)-1\right)(1-z)^2}dz $$
$$ =\int_0^1 \frac{z^3}{\left(\exp\left(\frac{z}{1-z}\right)-1\right)(1-z)^5}dz $$
Therefore
$$ f(z) = \frac{z^3}{\left(\exp\left(\frac{z}{1-z}\right)-1\right)(1-z)^5} $$
