Integration by Partial Fractions

1. Mar 4, 2014

cathy

1. The problem statement, all variables and given/known data
(2x^3-2x+1)/(x^2/3x)
Find the integral.

2. The attempt at a solution

So I've been on this problem for like an hour now and I don't know what I'm doing wrong. So I used long division and got
2x+ (4x+1)/(x^2-3x)

∫2x + ∫(4x+1)/(x^2-3x)
= x^2 + ∫(4x+1)/(x^2-3x).
For that next part, I used partial fractions.
A/x + B/(x-3) = 4x+1
Solved for a= -1/3
Solved for b= 13/3

Put everything together and solving for that integral, I got

x^2- 1/3ln(x) + 13/3ln(x-3)

and it's wrong.
Where am I going wrong? I'm so confused.
Thank for any help in advance.

2. Mar 4, 2014

vanhees71

Obviously there went something wrong with your partial-fraction calculation. I'd just do the following
$$\frac{2x^3-2x+1}{x^2-3x}=\frac{2x(x^2-3x)+6(x^2-3x)+16x+1}{x^2-3x}=2x+6 + \frac{16x+1}{x^2-3x}.$$
The first two terms are easily integrated. For the 2nd one you can find the partial fractions with the usual ansatz easily.

3. Mar 4, 2014

cathy

But i was told to use long division first, which was why I started with 2x+ (4x+1)/(x^2-3x).
And from there, I am confused, as to how I got the answer part wrong.

4. Mar 4, 2014

vanhees71

What I did is nothing else then long division (written in a way I can achieve with LaTeX).

5. Mar 4, 2014

HallsofIvy

Staff Emeritus
What vanhees71 is trying to tell you is that your long division is wrong. You should have "2x+ 6" as the quotient, not just "2x". And the remainder is "16x+ 1", not "4x+ 1".

I suspect you did not write the "$0x^2$" in the dividend and then mistook the "$-6x^2$" in the division for "$-6x$".

6. Mar 4, 2014

Staff: Mentor

The above has a typo. I'm pretty sure you meant the integrand to be (2x3 - 2x + 1)/(x2 - 3x)

7. Mar 4, 2014

cathy

Ohh, such a simple mistake with the long division. I was forgetting to write the variables I didn't use with 0. Thank you so much everyone.