(it doesn't make sense to have x in the limits when it is the variable you're integrating with respect to...EDIT, if it is pi, then no worries )
the way to handle an integral like this, is to simply expand the numerator, using the binomial theorem if n is large enough to make it useful, and then dividing each term by x. for the first term, the 1/x, the integral would be ln|x|, for the others, use the power rule for integrals: $\displaystyle \displaystyle \int x^n ~dx = \frac {x^{n + 1}}{n + 1} + C$
then apply the fundamental theorem of calculus to deal with the limits on the integral.
It's hard to read what the upper limit is, but let's supose it is pi, so:
$\displaystyle \displaystyle{\int\limits^\pi_0\frac{(1-x)^n}{x}\,dx=\int\limits^\pi_0\frac{\sum\limits^n_ {k=0}\binom{n}{k}(-x)^k}{x}\,dx=\int\limits^\pi_0\left(\frac{1}{x}-n+\frac{n(n-1)}{2}x-\ldots\pm x^{n-1}\right)\,dx$ ,
and now use linearity of the integral to solve each of the above...
Tonio
$\displaystyle \int \frac{(1-x)^n}{x} ~dx $ where $\displaystyle n $ is natural number .
Sub. $\displaystyle x = 1 -t $
$\displaystyle dx = -dt $
We have $\displaystyle - \int \frac{t^n}{1-t}~dt $
$\displaystyle = - \int \frac{ 1 - (1-t^n) }{1-t}~dt $
$\displaystyle = - \int \left[ \frac{1}{1-t} - (1+ t + t^2 + ... + t^{n-1} ) \right] ~dt $
$\displaystyle = \frac{t}{1} + \frac{t^2}{2} + \frac{t^3}{3} + .... + \frac{t^n}{n} + \ln|1-t| + C $
or $\displaystyle = \frac{(1-x)}{1} + \frac{(1-x)^2}{2} + \frac{(1-x)^3}{3} + ... + \frac{(1-x)^n}{n} + \ln|x| + C $