Integrating Rational Functions: Solving ∫(x/(x-1)^3)

whatlifeforme
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Homework Statement


integrate the following:

Homework Equations


∫(x/(x-1)^3

The Attempt at a Solution


i've tried u-substitution, finding an inverse trig function that matched the formula, and still can't figure out how to solve this problem.

u-subtitution for u=x gives the same problem. u-subsitution for x-1 gives du =1 which does not match the problem.
 
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whatlifeforme said:

Homework Statement


integrate the following:


Homework Equations


∫(x/(x-1)^3


The Attempt at a Solution


i've tried u-substitution, finding an inverse trig function that matched the formula, and still can't figure out how to solve this problem.

u-subtitution for u=x gives the same problem. u-subsitution for x-1 gives du =1 which does not match the problem.

u=(x-1) gives du=dx which does match the problem. If you are worried about the x in the numerator, if u=x-1, then x=u+1.
 
Its a fairly easy integral..dont go all complicated when you can't find the answer just stay on the ground bro :-p sometimes few problems can be easily solved if you just view it from a different angle

Dick said:
u=x-1, then x=u+1.

this should work. you would get u+1/u^3 du which you split and integrate
 
Hysteria X said:
Its a fairly easy integral..dont go all complicated when you can't find the answer just stay on the ground bro :-p sometimes few problems can be easily solved if you just view it from a different angle



this should work. you would get u+1/u^3 du which you split and integrate

No, you will not get u + \frac{1}{u^3}, which is what you wrote! If you really mean \frac{u+1}{u^3}, use parentheses, like this: (u+1)/u^3.
 
Hysteria X said:
this should work. you would get u+1/u^3 du which you split and integrate

Good advice but use more parentheses. You could easily mistake u+1/u^3 for u+(1/u^3) when you meant (u+1)/u^3.
 
Ray Vickson said:
No, you will not get u + \frac{1}{u^3}, which is what you wrote! If you really mean \frac{u+1}{u^3}, use parentheses, like this: (u+1)/u^3.

lol thanks for pointing it out..my bad :shy:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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