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Homework Help: Find the integral Please HELP

  1. Jan 26, 2008 #1
    Find the integral...Please HELP!!!!

    1. The problem statement, all variables and given/known data
    Find the integral...
    2t / (16t^4 + 1)


    2. Relevant equations



    3. The attempt at a solution
    I am stuck on this problem...I have tried long division but that didn't work out for me. I don't know how to solve this.
     
  2. jcsd
  3. Jan 26, 2008 #2
    Long division should be reserved for improper fractions.

    [tex]\int\frac{2t}{(4t^2)^2+1}[/tex]

    Can you take it from here?
     
  4. Jan 26, 2008 #3
    oh ok so i can use u substitution
    u = 4t^2
    du = 8t dt
    1/4du = 2t
    1/4 ln (4t^2)^2 + 1
    is that right?
     
  5. Jan 26, 2008 #4
    You're substitutions were right. But that is not correct.

    [tex]\frac 1 4 \int \frac{du}{u^2+1}[/tex]
     
  6. Jan 26, 2008 #5
    On my paper it says hint: use inverse trig functions. But I don't know how to use one of them because they don't resemble that.
     
  7. Jan 26, 2008 #6
    Keep searching! It's very common! Sine, Cosine, or Tangent.
     
  8. Jan 26, 2008 #7
    I can't use arctan because on the bottom it is a^2 + u^2 and u has to have the x in it right?
     
  9. Jan 26, 2008 #8
    It doesn't matter! You used a substitution and swapped variables.

    a, is simply a constant.

    [tex]\int\frac{dx}{x^2+a^2}=\int\frac{du}{u^2+1^2}[/tex]
     
  10. Jan 26, 2008 #9
    oh ok so I have...
    1/4 *1/1 arctan |4t^2| / 1 + C
    1/4 Arctan |4t^2| + C
     
  11. Jan 26, 2008 #10
  12. Jan 26, 2008 #11
    oh yay thanks!
    One more question...
    how do you find the integral of something with dx on top like...
    dx / (x^2 - 4x + 20)
     
  13. Jan 26, 2008 #12
    It doesn't matter what variables you're using, just pay attention to what you're Integrating with respects to and treat the others as constants.

    [tex]\int\frac{dx}{x^2-4x+20}[/tex]

    Try completing the square.
     
    Last edited: Jan 26, 2008
  14. Jan 26, 2008 #13
    I get...
    (x^2 - 4x + 4) - 4 + 20
    (x - 2)^2 + 16
     
  15. Jan 26, 2008 #14
    What does this Integral look like?

    [tex]\int\frac{dx}{(x-2)^2+16}[/tex]

    If you don't plan on using an Integral table, factor out a 16 from the denominator b/c you need a constant of 1.
     
  16. Jan 26, 2008 #15
    how do i factor out 16?? and why do I need a constant of 1?? I am so confused and I don't know why...I did all of the homework just fine.
     
  17. Jan 26, 2008 #16
    B/c before you can integrate this integral, you need to have it in the form similar to ...

    [tex]\int\frac{dx}{x^2+1}[/tex]

    Rather than factoring out a 16, divide both numerator & denominator by 16.
     
    Last edited: Jan 26, 2008
  18. Jan 26, 2008 #17
    Can anyone help me with this problem...I have a test Thursday and I am freaking out because I thought I understood this stuff!!
     
  19. Jan 27, 2008 #18
    [tex]\int\frac{dx}{(x-2)^2+16}[/tex]

    To factor out a 16 ...

    [tex]\frac{1}{16}\int\frac{dx}{\frac{(x-2)^2}{16}+1}[/tex]

    [tex]\frac{1}{16}\int\frac{dx}{\left(\frac{x-2}{4}\right)^2+1}[/tex]
     
    Last edited: Jan 27, 2008
  20. Jan 27, 2008 #19
    Is there any other way to solve this problem? I am only asking because we have never done anything like that in class or on any homework and I find it hard to believe our teacher giving us something we haven't done before. So I was just curious if there was another solution?
     
  21. Jan 27, 2008 #20
    It's no different from any of your other problems? You should be able to handle problems that are not 100% similar to what you have done b4. It's basically the integral of arctan.

    Completing the square and factoring is something you learned prior to Calculus, so those 2 initial steps should not bother you.
     
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