Integrating Factor Differential Equation

[Deathfish]

Homework Statement

e∫^P(x)

∫$\frac{x-2}{x(x-1)}$dx

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...

 

[SammyS]

Homework Statement

e∫^P(x)

∫$\frac{x-2}{x(x-1)}$dx

...

Need help...

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