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

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 | ?

Hurkyl
Staff Emeritus
Gold Member
What happened to the constant of integration?

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

Hurkyl
Staff Emeritus
Gold Member
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!

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

Hurkyl
Staff Emeritus