# How do I integrate this?

1. Nov 24, 2007

$$\int\frac{3x^4+4x^3+16x^2+20x+9}{x^5+2x^4+6x^3+12x^2+9x+18}dx$$

I tried using du/u for Lnu but I couldn't get it into that form. I don't have any idea what to do next, really. Can someone point me in the right direction?

2. Nov 25, 2007

### Ben Niehoff

You should try expanding it via partial fractions. There should be a section in your textbook about this.

3. Nov 25, 2007

### HallsofIvy

Staff Emeritus
I would have sworn that denominator couldn't be factored, but it can!

4. Nov 25, 2007

### Nachore

take u = (x^5+2x^4+6x^3+12x^2+9x+18)
du = 5x^4+8x^3+18x^2+24x+9
then take the numerator & cancel it out with du.

5. Nov 25, 2007

What?

I'm having trouble factoring it, and I'm not sure what Nachore means.

6. Nov 25, 2007

### coomast

Use Kramer on the denominator, you will find that -2 is a root.
The remaining part can be factorized easily.
Then use partial fractions to split the integral into more classical ones.

This is not a particularly difficult integral, although a very tedious one to solve.
You might say that this makes it a difficult one after all.

Edit: It is not as tedious as I thought at first, the solution is ln(x+2)+ln(x^2+3)-2/(x^2+3)+C

Last edited: Nov 25, 2007
7. Nov 25, 2007

### HallsofIvy

Staff Emeritus
Yes, that's true. but since the numerator is nothing like that, it doesn't help at all!

8. Nov 27, 2007

I get (x+2)(x^2+3)^2 as the bottom, factored out. Is this correct? I'm really not good at factoring.

9. Nov 27, 2007

### HallsofIvy

Staff Emeritus
Yes, that's correct. Now use partial fractions.