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Tricky Exponential/Logarithmic Indefinite Integrals

  1. Oct 4, 2009 #1
    Hi,
    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

    l_9808c71c1e404b6f9d82cdd0d043e4e2.jpg

    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

    l_30b304abd77d4d16ad436c6b08542e24.jpg

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

    l_3bfaa8ed9e5f413dba40e7ffdf7e13ab.jpg
     
  2. jcsd
  3. Oct 5, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    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
    [tex]
    \frac{1}{\sqrt{2+ u^3}}
    [/tex]
    anywhere...or at least I shouldn't!
     
  5. Oct 5, 2009 #4

    Dick

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    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.
     
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