anonymous
  • anonymous
evaluate the integral 5/(n(n+9)) from 1 to infinity. Please Help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
can anyone help????
anonymous
  • anonymous
maybe partial fractions, but I need help on that part too...
anonymous
  • anonymous
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: \[\lim_{b \rightarrow \infty}\int\limits_{1}^{b}[5/(9n)-5/(9(n + 9))]dn\]

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anonymous
  • anonymous
Awww yeah, thats how it's done
anonymous
  • anonymous
Just use some u substitution and your done!
anonymous
  • anonymous
Other than the limit, just use log rules to combine them and it should work out
anonymous
  • anonymous
Factor out a 5/9 to make the calculations somewhat easier: \[(5/9)\int\limits_{1}^{b}[1/n-1/(n+9)]dn\] \[(5/9)[\ln |n|-\ln |n+9|]_{1}^{b}\] \[(5/9)[\ln |b|-\ln |b+9|-\ln1+\ln10]\] \[(5/9)\ln[10|b/(b+9)|]\]
anonymous
  • anonymous
Take the limit as b approaches infinity of the last thing, I got (5/9)*ln10
anonymous
  • anonymous
I think it's e^ of that answer. I don't think you can use L'hospitals rule inside logs, can you?
anonymous
  • anonymous
I didn't use L'Hopital's rule. I divided the top and the bottom of the fraction by b.
anonymous
  • anonymous
yeah... it's just (5/9)ln10
anonymous
  • anonymous
thank you so much, that was so hard for me!
anonymous
  • anonymous
Try wolframalpha.com for these things, if you just want a solution, by the way.
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=integrate%205%2F(n(n%2B9))%20from%201%20to%20infinity&t=macw01

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