The equation is given below where $z = a+ bx$
$\int { dx \over {x^n z^m}} = {{-1 \over{(n-1)ax^{n-1}z^{m-1}}} + {b(2-n-m) \over a(n-1)} \int {dx \over x^{n-1}z^m} }$
Integration by parts gives:
$(a+bx)^{-m}\int x^{-n}dx - \int ({d \over dx}(a+bx)^{-m} \int x^{-n} dx)dx $
$= {1 \over{-(n-1)(a+bx)^m x^{n-1}} }- {mb \over (n-1)} \int {dx \over (a+bx)^{m+1}x^{n-1}}$
Now the problem is I can't match the powers of my equation with the given RHS. Either x or z has a different order from which I found! How can I do that right? Thanks
Idea, too long for a comment: $$ \int{dx\over{x^n z^m}} = {{-1 \over{(n-1)ax^{n-1}z^{m-1}}} + {b(2-n-m)\over a(n-1)}\int{dx\over x^{n-1}z^m} } $$ is equivalent to $$ {1\over{x^n z^m}} = \frac{d}{dx}{{-1 \over{(n-1)ax^{n-1}z^{m-1}}} + {b(2-n-m)\over a(n-1)}{1\over x^{n-1}z^m}}, $$ and this must be easy to check.
EDIT:
Your formula is wrong: when $a =1$, $b = 0$: $$ \frac{d}{dx}{{-1\over{(n-1)ax^{n-1}z^{m-1}}}} = \frac{d}{dx}{{-1\over{(n-1)x^{m+n-2}}}} = \frac{m + n - 2}{(n - 1)x^{m+n-1}}, $$ while: $$ {1\over{x^n z^m}} - {b(2-n-m)\over a(n-1)}{1\over x^{n-1}z^m} = {1\over{x^{m+n}}}. $$