# Integrating Factor Differential Equation

1. Feb 28, 2012

### Deathfish

1. The problem statement, all variables and given/known data

e∫^P(x)

∫$\frac{x-2}{x(x-1)}$dx

3. The attempt at a solution

so i split it into

∫$\frac{x-2}{x(x-1)}$dx
= ∫$\frac{2x-1}{x^2-x}$dx - ∫$\frac{x+1}{x^2-x}$dx

= ln(x2-x) - ∫$\frac{x}{x^2-x}$ - ∫(x2-x)-1

= ln(x2-x) - ln(x-1) - ∫(x2-x)-1

ok. having problems working out ∫(x2-x)-1dx
tried many ways but i keep ending up with the original integral.

u=x-1 --> du=-x-2
dv= (x-1)-1dx --> v=ln(x-1)

gives me ($\frac{1}{x}$)ln(x-1) + ∫$\frac{ln(x-1)}{x^2}$dx

when i work this out

∫$\frac{ln(x-1)}{x^2}$dx

u=ln(x-1) --> du=$\frac{1}{x-1}$
dv=x-2dx --> v=-x-1

i get

∫$\frac{ln(x-1)}{x^2}$dx = $\frac{-ln(x-1)}{x}$ + ∫(x2-x)-1

which is the same integral and above and i get no solution.
Need help......

2. Feb 28, 2012

### SammyS

Staff Emeritus
Use partial fraction decomposition on $\displaystyle\frac{x-2}{x(x-1)}\,.$