Find the integral of x/(x-1)^2 with respect to x

[Beretta]

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 happened to the x in ->>> x ( -1 / (x-1) ) + ln | x - 1 | ?

Thank you in advance

 

[Hurkyl]

What happened to the constant of integration?

 

[Beretta]

I'm using this in differential equations I don't need the constant now.

 

[Hurkyl]

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]

May you show me how I can elliminate the x please?

 

[Hurkyl]

Subtract your answer from the "right" answer: what's left?

 

[Beretta]

thank you SO much!