Stuck on an integral

  • Thread starter duelle
  • Start date
  • #1
3
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1. The problem
Find the definite integral.
[tex]\int \frac {t^2 - 3}{-t^3 + 9t + 1} dt[/tex]


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

Answers and Replies

  • #2
22
0
yes that rule is used, but when you take du, you're entire quantity needs to be multiplied by dt.
 
  • #3
682
1
the last line of your calculation is wrong.

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

you forgot changing the dt.
 
  • #4
4
0
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:
  • #5
3
0
Wow, I feel dumb. Thanks for pointing that out, though.
 

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