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anonymous

  • 5 years ago

OK, I can't figure this improper integral out -- I keep getting -ln2 and the answer is supposed to be ln2; integrate from -1 to infinity dx/(x^2+5x+6); I've tried partial fractions -- but that doesn't seem to work because I get 1/(x+2) + 0/(x+3) . . . and I can't do anything with that! Help!

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  1. anonymous
    • 5 years ago
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    For the partial fractions, did you get 1=A(x+2)+B(x+3). This works out.

  2. anonymous
    • 5 years ago
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    Well, I'll try again . . . I know I'm tired . . . so I'll try again!

  3. anonymous
    • 5 years ago
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    SO . . . this time I got 1/(x+2) and -1/(x+3) . . . ?? that would mean that 1/(x^2+5x+6) = 1/((x+2) -1/((x+3) . . . am I on the right track?

  4. anonymous
    • 5 years ago
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    Thanks for your help . . . I'll keep working. Have a good evening.

  5. anonymous
    • 5 years ago
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    Thats what i got for the decomposition. Good luck

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