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Integration through substitution help please

cheetahday

New member
Joined
Jan 14, 2015
Messages
1
\(\displaystyle \displaystyle{ \int \,}\)\(\displaystyle \dfrac{dx}{x\, \left(x^p\, +\, a\right)},\) 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
 
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Ishuda

Elite Member
Joined
Jul 30, 2014
Messages
3,345
\(\displaystyle \displaystyle{ \int \,}\)\(\displaystyle \dfrac{dx}{x\, \left(x^p\, +\, a\right)},\) 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

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