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

Question

anonymous · Answer

can anyone help????

anonymous · Answer

maybe partial fractions, but I need help on that part too...

anonymous · Answer

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:

$$\lim_{b \rightarrow \infty}\int\limits_{1}^{b}[5/(9n)-5/(9(n + 9))]dn$$

anonymous · Answer

Awww yeah, thats how it's done

anonymous · Answer

Just use some u substitution and your done!

anonymous · Answer

Other than the limit, just use log rules to combine them and it should work out

anonymous · Answer

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 · Answer

Take the limit as b approaches infinity of the last thing, I got (5/9)*ln10

anonymous · Answer

I think it's e^ of that answer. I don't think you can use L'hospitals rule inside logs, can you?

anonymous · Answer

I didn't use L'Hopital's rule.  I divided the top and the bottom of the fraction by b.

anonymous · Answer

yeah... it's just (5/9)ln10

anonymous · Answer

thank you so much, that was so hard for me!

anonymous · Answer

Try wolframalpha.com for these things, if you just want a solution, by the way.

anonymous · Answer

http://www.wolframalpha.com/input/?i=integrate%205%2F(n(n%2B9))%20from%201%20to%20infinity&t=macw01