# Integration by partial fractions

1. Jan 22, 2006

### Xcron

I started this section off quite well and I did very well on the problems where there are only linear factors but when I got to the problems with quadratic factors, I began getting wrong answers. I though that perhaps I would receive some advice or my error/mistake could be corrected if possible.

The problem is:
$$\int\frac{4x^4-43x^3+200x^2-442x+383}{(x-3)(x^2-6x+13)}dx$$

It comes with a small help thing that says: $$(x-3)(x^2-6x+13) = x^3-9x^2+31x-39$$

Ok so the first thing I did was use long division and I got $$4x-7$$ as the answer of the long division. Next I proceeded to do:
$$4x^4-43x^3+200x^2-442x+383 = A(x^2-6x+13) + (Cx+D)(x-3)$$.
Next, I plugged in 3 for the value of x so that I could solve for A and I got its value to be 5. Next, I plugged in 0 for the value of x so that I could get A and D alone and solve for D. I got its value to be -106, which is horribly wrong for some reason, I think. Next I plugged in 1 for the value of x and solved for C nd its value came out to be 75, which is also horribly wrong probably..I made a mistake somewhere in there I think...after that, I set up the new integral:
[/tex]\int\(4x-7+\frac{5}{x-3}+\frac{75x-106}{x^2-6x+13})dx[/tex].
I solved that and got my final answer to be:
$$2x^2-7x+5\ln|x-3|+\frac{75}{2}\ln|x^2-6x+13|+ \frac{119}{2}\arctan(\frac{x-3}{2})+C$$.

$$2x^2-7x+5\ln|x-3|+4\ln|x^2-6x+13|+ \frac{9}{2}\arctan(\frac{x-3}{2})+C$$.

If I am making some kind of mistake, then I must be inherently making it because all my other answers to problems with quadratic factors is coming out wrong...please help me.

2. Jan 22, 2006

### durt

You should have a remainder after you divide:
$$4 x - 7 + \frac{13 x^2 - 69 x + 10}{x^3 - 9 x + 31 - 39}$$
That's an awfully tedious problem. :yuck:

3. Jan 22, 2006

### Xcron

Yes, I do have a remainder after I divide. After the long division, I have:
$$4 x - 7 + \frac{13 x^2 - 69 x + 110}{x^3 - 9 x^2 + 31x - 39}$$

What do I do with it? I was just looking over that because I noticed something about it in the book, but I can't quite grasp what the remainder would be used for..

Last edited: Jan 22, 2006
4. Jan 22, 2006

### durt

You'll want to split the remainder into partial fractions and then you can integrate everything.

5. Jan 22, 2006

### Xcron

Ok, I just tried that and I finally got the right answer...I guess that's what my error was. After the long division, I need to use the remainder..heh..

Last edited: Jan 22, 2006
6. Jan 22, 2006

### Xcron

Upon further analysis, I realized that the answer is still wrong, this is why:

after the division and the $$a^2 + x^2$$ trig substitution, the integral is:
$$2x^2-7x+5\ln|x-3|+ \int\frac{16\tan\theta+24-15}{4\sec^2\theta}2\sec^2\theta d\theta$$
then:
$$x^2-7x+5\ln|x-3|+8\ln|sec\theta|+\frac{9}{2}\theta$$

The 8 in front of the natural log is supposed to be a four, anyone know why I have an 8?...