cheetahday

New member
$$\displaystyle \displaystyle{ \int \,}$$$$\displaystyle \dfrac{dx}{x\, \left(x^p\, +\, a\right)},$$ 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
$$\displaystyle \displaystyle{ \int \,}$$$$\displaystyle \dfrac{dx}{x\, \left(x^p\, +\, a\right)},$$ 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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