1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How do you integrate (a^x + b^x)^3 /((a^x)*(b^x))

  1. May 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Integrate (a^x + b^x)^3 /((a^x)*(b^x))


    3. The attempt at a solution

    I have only 2 semesters of calculus under my belt yet nothing in my experience has taught me to do anything like this
     
  2. jcsd
  3. May 20, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi The_ArtofScience! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    erm :redface: … just expand it …

    (a2/b)x + … :smile:
     
  4. May 21, 2009 #3
    The algebra that I did from expanding it like you suggested was:

    (a^3x + 3a^2x b^x + 3a^x b^2x + b^3x)/(a^x b^x)
    = (a^2 /b)^x + 3a^x + 3b^x + (b^2/a)^x

    But then what do I do with this strange exponent? I don't know how to integrate an expression like b^x
     
  5. May 21, 2009 #4

    Born2bwire

    User Avatar
    Science Advisor
    Gold Member

    You can find it easily enough by a search but you can derive it pretty easily too.

    [tex]\log_b(b^x) = x[/tex]
    [tex]\frac{d}{dx}\log_b(b^x) = 1[/tex]
    [tex]\frac{d}{dx}\log_b(b^x) = \frac{d}{du} \log_b(u) \cdot \frac{d b^x}{dx}[/tex] where [tex]u = b^x[/tex].

    Convert [tex]log_b(u) = \frac{\ln(u)}{\ln(b)}[/tex] and go from there.
     
  6. May 21, 2009 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    (please use the X2 tag just above the Reply box … it's much easier to read)

    Yes, use Born2bwire's :smile: method,

    or write bx = (eln(b))x = eln(b)x,

    so (bx)' = ln(b)bx :wink:
     
  7. May 21, 2009 #6

    Born2bwire

    User Avatar
    Science Advisor
    Gold Member

    Oh sure, do it the easy and efficient way. ;)
     
  8. May 22, 2009 #7
    Hi I just read the messages and I appreciate both of your help. Thanks! I get it now. It was silly for me not to see it before I asked
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook