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anonymous

  • 5 years ago

evaluate the integral 5/(n(n+9)) from 1 to infinity. Please Help

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  1. anonymous
    • 5 years ago
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    can anyone help????

  2. anonymous
    • 5 years ago
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    maybe partial fractions, but I need help on that part too...

  3. anonymous
    • 5 years ago
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    First, you should break up the 5/(n(n + 9)) into two separate fractions. A/n + B/(n + 9) = 5/(n(n+9)) --> A(n + 9) + Bn = 5 Let n = 0 to find A = 5/9 and n = -9 to find B = -5/9. You should find that 5/(n(n + 9)) = 5/(9n) - 5(9(n + 9)). You should be wanting to integrate this: \[\lim_{b \rightarrow \infty}\int\limits_{1}^{b}[5/(9n)-5/(9(n + 9))]dn\]

  4. anonymous
    • 5 years ago
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    Awww yeah, thats how it's done

  5. anonymous
    • 5 years ago
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    Just use some u substitution and your done!

  6. anonymous
    • 5 years ago
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    Other than the limit, just use log rules to combine them and it should work out

  7. anonymous
    • 5 years ago
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    Factor out a 5/9 to make the calculations somewhat easier: \[(5/9)\int\limits_{1}^{b}[1/n-1/(n+9)]dn\] \[(5/9)[\ln |n|-\ln |n+9|]_{1}^{b}\] \[(5/9)[\ln |b|-\ln |b+9|-\ln1+\ln10]\] \[(5/9)\ln[10|b/(b+9)|]\]

  8. anonymous
    • 5 years ago
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    Take the limit as b approaches infinity of the last thing, I got (5/9)*ln10

  9. anonymous
    • 5 years ago
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    I think it's e^ of that answer. I don't think you can use L'hospitals rule inside logs, can you?

  10. anonymous
    • 5 years ago
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    I didn't use L'Hopital's rule. I divided the top and the bottom of the fraction by b.

  11. anonymous
    • 5 years ago
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    yeah... it's just (5/9)ln10

  12. anonymous
    • 5 years ago
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    thank you so much, that was so hard for me!

  13. anonymous
    • 5 years ago
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    Try wolframalpha.com for these things, if you just want a solution, by the way.

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