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!

Homework Help: Tricky Exponential/Logarithmic Indefinite Integrals

  1. Oct 4, 2009 #1
    Our professor has only taught us these methods for Integration...thus far:
    Direct Integration
    Substituion Method

    So theoretically we should be able to solve these problems by substitution, direct integration...(double substitution) & by algebraic manipulation etc.

    Ive attached 3 attempts at answers for 3 problems:

    This first problem I am curious how our professor got "the answer in question"...Ive tried my best but all I could get was "my answer" the logic seems correct...see image


    This second problem I am curious why our professor also gave "the answer in question"...Ive tried my best but I would like to know if I am solving this correctly "my answer"...see image


    This last problem I am just curious to see if this is the correct way to solve this problem "my answer"...see image

  2. jcsd
  3. Oct 5, 2009 #2


    User Avatar
    Science Advisor

    Your difficulty appears to be bad algebra. In the first problem you have
    [tex]\frac{2e^{2x}- 8/3}{2e^{2x}- 5}[/tex]
    and cancel the "[itex]2e^{2x}[/itex]" terms. You can't do that.

    In the second problem you have, in one equation,
    [tex]\frac{1}{\sqrt{2+ u^3}}[/itex]
    and in the next, the square root has magically disappeared!

    In the third problem, you have
    [tex]\frac{1}{2}\int 9^{u^2} du[/tex]
    and somehow that becomes
    [tex]\frac{1}{2}\int log_9(u)du[/itex]
    How did the exponential become a logarithm?
  4. Oct 5, 2009 #3
    Thanks for the reply and suggestions...

    Ive since gotten the answers to problems #1(didnt balance the algebra correctly) and #3 (used the wrong integral formula)..

    However, I believe #2 is correct...perhaps my posted image is bad...I dont have the equation
    \frac{1}{\sqrt{2+ u^3}}
    anywhere...or at least I shouldn't!
  5. Oct 5, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper

    In #2 after you do the substitution you have 1/sqrt(1+u^2). But the integration formula you apply is for 1/(a^2+u^2) (without the square root). You need an integration formula for 1/sqrt(u^2+a^2).
  6. Oct 5, 2009 #5
    In problem 2, the obvious and natural method to evaluate [tex]\int\frac{1}{\sqrt{1+u^2}}\,du[/tex] is trig substitution. Unfortunately, you said you are not allowed to use that method.

    You could use the completely unmotivated trick of multiplying numerator and denominator by [tex]u+\sqrt{1+u^2}[/tex]. Then the substitution [tex]w=u+\sqrt{1+u^2}[/tex] works.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook