1. Sep 22, 2007

frasifrasi

Can anyone confirm that the answer
for int (1/x(x-1))dx

is

- ln|x|+ ln |x-1| + c using partial fractions?

Thank you.

2. Sep 22, 2007

Staff: Mentor

How about expanding $$\frac{1}{x (x-1)}$$ into

$$\frac{A}{x}\,+\,\frac{B}{x-1}$$ and solve for A and B

then solve the integral

3. Sep 22, 2007

HallsofIvy

Staff Emeritus
Can't you "confirm" it yourself by differentiating? The derivative of -ln |x|+ ln|x-1|+ c
(assuming for the moment that x> 1 so both x and x-1 are positive) is -1/x+ 1/(x-1)= -(x-1)/(x(x-1))+ x/(x(x-1))= 1(x(x-1)). (You might want to combine those two logarithms.)

I suspect that is exactly what he did! Doing it again is not a "check".