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)}\,.