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Beretta
Oct16-04, 06:38 PM
Integral (x / (x-1)^2) dx =

Let u = x, du = dx
Let dv = 1 / (x-1)^2 , v = -1 / x-1

uv - integral (vdu) dv

x ( -1 / (x-1) ) + integral (1 / x-1 ) =

x ( -1 / (x-1) ) + ln | x - 1 |

how come the answer is ( -1 / (x-1) ) + ln | x - 1 | ? what happend to the x in ->>> x ( -1 / (x-1) ) + ln | x - 1 | ?

Thank you in advance

Hurkyl
Oct16-04, 06:43 PM
What happened to the constant of integration?

Beretta
Oct16-04, 06:48 PM
I'm using this in differential equations I don't need the constant now.

Hurkyl
Oct16-04, 07:16 PM
Ah, but an indefinite integral is a multivalued operation; you always need the constant of integration!

It turns out that this is problem is a good example of why it matters! When you account for the constant of integration, you can show that your answer and the "right" answer are the same!

Beretta
Oct16-04, 07:24 PM
May you show me how I can elliminate the x please?

Hurkyl
Oct16-04, 07:25 PM
Subtract your answer from the "right" answer: what's left?

Beretta
Oct16-04, 07:31 PM
thank you SO much!