Integration through substitution help please

cheetahday

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\displaystyle \displaystyle{ \int \,}dxx(xp+a),\displaystyle \dfrac{dx}{x\, \left(x^p\, +\, a\right)}, where a,p0.\displaystyle a,\, p\, \neq\, 0. Use the substitution u=1+axp.\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
 
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\displaystyle \displaystyle{ \int \,}dxx(xp+a),\displaystyle \dfrac{dx}{x\, \left(x^p\, +\, a\right)}, where a,p0.\displaystyle a,\, p\, \neq\, 0. Use the substitution u=1+axp.\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

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

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

or

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

and

dxx(xp+a)=(pa)1xp+1duxp+1(1+axp)=(pa)1duu\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}
 
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