Integration through substitution help please

[cheetahday]

$\displaystyle \displaystyle{ \int \,}$$\displaystyle \dfrac{dx}{x\, \left(x^p\, +\, a\right)},$ where $\displaystyle a,\, p\, \neq\, 0.$ Use the substitution $\displaystyle u\, =\, 1\, +\, ax^{-p}.$

So far I've only gotten
x^p=(u-1)/a
x=((u-1)/a)^(1/p)

I'm stuck everything I plug in just becomes really messy

 

[Ishuda]

$\displaystyle \displaystyle{ \int \,}$$\displaystyle \dfrac{dx}{x\, \left(x^p\, +\, a\right)},$ where $\displaystyle a,\, p\, \neq\, 0.$ Use the substitution $\displaystyle u\, =\, 1\, +\, ax^{-p}.$

So far I've only gotten
x^p=(u-1)/a
x=((u-1)/a)^(1/p)

I'm stuck everything I plug in just becomes really messy

$\displaystyle u\, =\, 1\, +\, a\, x^{-p}$

$\displaystyle du\, =\, -pa\, x^{-(p\, +\, 1)}\, dx$

or

$\displaystyle dx\, =\, -(pa)^{-1}\, x^{p\, +\,1}$

and

$\displaystyle \dfrac{dx}{x\left(x^p \,+\, a\right)} \,=\, \dfrac{-(pa)^{-1} \,x^{p+1} \,du}{x^{p+1}\left(1\, +\, a x^{-p}\right)}\,=\, \dfrac{-(pa)^{-1}\, du}{u}$