# Stuck on an integral

1. Mar 26, 2007

### duelle

1. The problem
Find the definite integral.
$$\int \frac {t^2 - 3}{-t^3 + 9t + 1} dt$$

2. The attempt at a solution
The answer is the book made it seem like the rule $\int \frac {du}{u} = ln|u| + C$ was used. Here's what I got:
$$u = -t^3 + 9t + 1$$
$$du = -3t^2 + 9\;dt$$
$$-\frac{1}{3}du = t^2 + 9\;dt$$
Obviously this won't work out, so is there something I'm overlooking that anyone can point out?

Last edited: Mar 26, 2007
2. Mar 26, 2007

### a9211l

yes that rule is used, but when you take du, you're entire quantity needs to be multiplied by dt.

3. Mar 26, 2007

### tim_lou

the last line of your calculation is wrong.

notice that it should be:
$$-\frac{1}{3}du=(t^2-3)dt$$

you forgot changing the dt.

4. Mar 26, 2007

### Spiro09

you've got the right substitution. By rearranging your du formula for dt you'll get:

dt = -du/(3(t^2-3))
and it will cancel out leaving you with the simple integral

Last edited: Mar 26, 2007
5. Mar 27, 2007

### duelle

Wow, I feel dumb. Thanks for pointing that out, though.