# Stuck with an integral

1. Jan 20, 2013

### grimireaper

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

2. Relevant equations

I started with (2x^2+3)/(x^3-2x^2+3x) dx

3. The attempt at a solution

Got dx/x + (x+2)/(x^2-2x+3) dx using the A/x + (Bx+C)/(x^2-2x+3) method.

But I can't seem to solve (x+2)/(x^2-2x+3) dx

Possible solutions are welcome.

Last edited: Jan 20, 2013
2. Jan 20, 2013

### LCKurtz

Welcome to PF. You won't get solutions here, but you may get hints. Try this:$$\frac {x+2}{x^2-2x+3} = \frac{(x-1)+3}{(x-1)^2+2}$$and see if that gives you any ideas.

3. Jan 20, 2013

### grimireaper

I thought of that but the +2 at the bottom ruins it, because splitting up x-1 and 3 into 2 integrals won't help, as I cant cancel anything out

4. Jan 20, 2013

### LCKurtz

But you can break it up into two fractions and use different methods on them.

5. Jan 20, 2013

### lurflurf

$$2x^2+3=(x^2-x)+(3x)+(x^2-2x+3)=\frac{x}{2}\left(\frac{x^3-2x^2+3x}{x}\right)^\prime+3\frac{x^3-2x^2+3x}{x^2-2x+3}+\frac{x^3-2x^2+3x}{x}$$

6. Jan 20, 2013

### grimireaper

Got it solved. Thank you for the help.