Integration of a certain function Any suggestion on how to evaluate the following?
$$\int_a^b \Big(\frac{b-x}{x}\Big)^{\frac{1}{n-1}} dx$$ 
where $0<a<b<+\infty$ and $n\geq 2$ is an integer.
Any idea for substitution or integration by reduction? Preferably, a method that can also be useful in solving 
$$\int_a^b \Big(\frac{b-x}{x}\Big)^{\frac{n}{n-1}} dx.$$
 A: There is certainly a more knowledgeable way in terms of the incomplete Beta function, as pointed out by i707107. Here is nevertheless a naive approach that will take you to a closed form.
After the substitution
$$
u=\Big(\frac{b-x}{x} \Big)^\frac{1}{n-1}
$$
and an integration by parts, we end up computing
$$
\int \frac{1}{1+u^{n-1}}du.
$$
And the latter can be done by partial fraction decomposition.
The same steps, for the second integral, take you to
$$
\int \frac{u^{n-1}}{1+u^{n-1}}du=u-\int \frac{1}{1+u^{n-1}}du.
$$

Details: the substitution
$$
u=\Big(\frac{b-x}{x} \Big)^\frac{1}{n-1}\quad\iff\quad x=\frac{b}{1+u^{n-1}}
$$
is a $C^1$-diffeomorphism and yields
$$
b\int_0^{u_a}\frac{u\cdot (n-1)u^{n-2}}{(1+u^{n-1})^2}du\quad\mbox{where}\quad u_a=\Big(\frac{b-a}{a}\Big)^\frac{1}{n-1}.
$$
By integration by parts, we get
$$
-\frac{bu}{1+u^{n-1}}\Big|_0^{u_a}+b\int_0^{u_a}\frac{1}{1+u^{n-1}}du=-a\Big(\frac{b-a}{a}\Big)^\frac{1}{n-1}+b\int_0^{u_a}\frac{1}{1+u^{n-1}}du.
$$
Now we need to decompose the rational fraction into partial fractions. First
$$
u^{n-1}+1=\prod_{k=0}^{n-2}\big(u-\omega_k\big)\qquad\mbox{where}\quad \omega_k=e^{\frac{i(2k+1)\pi}{n-1}}
$$
has $n-1$ simple roots, whence
$$
\frac{1}{1+u^{n-1}}=\sum_{k=0}^{n-2}\frac{\alpha_k}{u-\omega_k}\qquad\mbox{with}\quad \alpha_k=\frac{1}{(n-1)\omega_k^{n-2}}=-\frac{\omega_k}{n-1}.
$$
It turns out that given our interval of integration and the $\omega_k$'s, we can use the principal branch of the complex logarithm to get
$$
\int_0^{u_a}\frac{1}{1+u^{n-1}}du=-\frac{1}{n-1}\sum_{k=0}^{n-2}\omega_k\log\Big(\frac{u_a-\omega_k}{-\omega_k} \Big) .
$$
Finally, we get

$$
\int_a^b\Big(\frac{b-x}{x} \Big)^\frac{1}{n-1}dx=-a\Big(\frac{b-a}{a}\Big)^\frac{1}{n-1}+\frac{b}{n-1}\sum_{k=0}^{n-2}\omega_k\log\Big(\frac{\omega_k}{\omega_k-u_a} \Big) .
$$

Check: for $n=2$, this does give what we should expect, namely $-(b-a)+b\log(b/a)$.
