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**Integral help!!**

## Homework Statement

Evaluate the following indenite integral,

∫[itex]\frac{x^3-5x}{x^3-2x^2+4x-8}[/itex] dx

## Homework Equations

I think I am right until i get to equating the equations when I do it the method im taught in class I get c=0 and I dont think that is right. However when i put it into a system of equations with the coefficients i get fractions.

I would like to know what to do next. Thanks

## The Attempt at a Solution

First the degree of numerator is same as the denomenator. So i do long division and end up with

∫1+ [itex]\frac{2x^2-9x+8}{x^3-2x^2+4x-8}[/itex]

Now decompose the fraction part into:

[itex]\frac{2x^2-9x+8}{(x^2+4)(x-2)}[/itex] =[itex]\frac{Ax+B}{(x^2+4)}[/itex]+[itex]\frac{C}{(x-2)}[/itex]

Then get a common denomenator for RHS

= [itex]\frac{(Ax+B)(x-2)+C(x^2+4)}{(x^2+4)(x-2)}[/itex]

From here i tried if x=2 to cancel out the (Ax+B) term which results in C=0

this is where i believe an error has occured.

My other thinking was can I expand then factor out the x^2 and x and constant leaving the A and B and C in whatever they factor into and then do a system of equations.

My result this way is

A= 9/4, B =-9/2, C=-1/4

Kind regards