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Integration exam question

  1. Aug 10, 2011 #1
    1. The problem statement, all variables and given/known data
    find the integral from 0 to 3 of (5x^2+30)/(x^4+13x^2+36)


    2. Relevant equations



    3. The attempt at a solution
    I thought it may be an integration by parts question, so I tried that first, but the book says that I should be using substitution.
    So I tried using u=x^2:
    du=2xdx
    x=u^(1/2)

    (5u+30)du/(u^2+13u+36)(2u^(1/2))
    so now I have
    (5u+30)du/(2u^(5/2)+26u^(3/2)+72u^(1/2))

    but i have no better idea how to integrate that than i did the original equation. any ideas?
     
  2. jcsd
  3. Aug 10, 2011 #2
    This problem would probably be best done by partial fraction decomposition first, then substitutions.
     
  4. Aug 10, 2011 #3

    Mark44

    Staff: Mentor

    I agree with Bohrok.
     
  5. Aug 10, 2011 #4
    do you mean using something like
    (Ax+B)/(x^2+4)+(Cx+D)/(x^2+9)=5x^2+30?
     
  6. Aug 10, 2011 #5

    Mark44

    Staff: Mentor

    Not quite.
    It would be

    (Ax+B)/(x^2+4) + (Cx+D)/(x^2+9) = (5x^2+30)/[(x^2+4)(x^2+9)]

    The goal is to solve for A, B, C, and D so that this equation is identically true.
     
  7. Aug 10, 2011 #6
    okay, so i solved and got
    A=0
    B=3
    C=0
    D=2

    so then i had
    3/(x^2+4)+2/(x^2+9)

    which i could integrate to get
    3ln(x^2+4)+2ln(x^2+9)

    and i could continue, but my possible answers all involve arctan so i think i'm on the wrong track here...?
     
  8. Aug 10, 2011 #7

    Mark44

    Staff: Mentor

    I didn't check your values for A, B, etc.

    Your integrals are incorrect. You should not be getting log expressions, but you should be getting arctan expressions.
     
  9. Aug 10, 2011 #8

    Mark44

    Staff: Mentor

    Also, check your work for the constants.
    3/(x^2 + 4) + 2/(x^2 + 9) = (5x^2 + 35)/[(x^2 + 4)(x^2 + 9)] [itex]\neq[/itex] (5x^2 + 30)/[(x^2 + 4)(x^2 + 9)]
     
  10. Aug 10, 2011 #9
    how do i get arctan expressions? because the derivative of tan is 1/(a^2+x^2), and i have 2 and 3 rather than 1 as my numerators?
     
  11. Aug 10, 2011 #10
    sorry..
    3/(x^2+9)+2/(x^2+4) ,right?
     
  12. Aug 10, 2011 #11

    Mark44

    Staff: Mentor

    No, d/dx(tan(x)) = sec2(x).

    d/dx(tan-1(x)) = 1/(x2 + 1) - that must be what you're thinking about.

    Also, d/dx(tan-1(x/a)) = (1/a)/((x/a)2 + 1)
    This implies that ∫(1/a)/((x/a)2 + 1) dx = tan-1(x/a) + C
     
  13. Aug 10, 2011 #12
    i'm confused....
     
  14. Aug 10, 2011 #13
    If d/dx (x2) = 2x, then ∫2x dx = x2 + C by integrating both sides. Mark just did the same thing with tan-1(x/a) where a is a constant.
     
  15. Aug 10, 2011 #14
    okay i think that makes sense... but how can i use that to solve the problem?
     
  16. Aug 10, 2011 #15
    That'll come in later; first check over your work for the constants in the numerators of the partial fractions. The numbers are almost correct.
     
  17. Aug 10, 2011 #16
    i had
    (ax+b)(x^2+4)+(cx+d)(x^2+9)=(5x^2+30)
    ax^3+bx^2+4Ax+4B+Cx^3+9Cx+Dx^2+9D=5X62+30

    A+C=0 and 4A+9C=0 therefore A and C both equal 0
    B+D=5 and 4B+9D=30, so 5D=10 and D=2 so B=3
    and i have
    3/(x^2+9)+2(x^2+4)

    :( can't figure out what's wrong!
     
  18. Aug 10, 2011 #17

    vela

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    Earlier, you wrote
    which wasn't correct. Now, you have
    which is correct. You swapped the coefficients earlier.
     
  19. Aug 10, 2011 #18

    Mark44

    Staff: Mentor

    This is correct. You can easily check by verifying that 3/(x^2+9)+2/(x^2+4) = (5x^2 + 30)/[(x^2+9)(x^2+4)]
     
    Last edited: Aug 10, 2011
  20. Aug 10, 2011 #19
    okay, so from here i should integrate and end up with arctan functions... how do i do this?
     
  21. Aug 10, 2011 #20

    Mark44

    Staff: Mentor

    [tex]\int \frac{a~dx}{x^2 + a^2} = tan^{-1}(x/a) + C[/tex]

    This formula is essentially the same as the one I gave in an earlier post.

    A mistake you made earlier was thinking that the following was true:
    [tex]\int \frac{dx}{x^2 + a^2} = ln(x^2 + a^2) + C[/tex]

    More generally, this also is not true:
    [tex]\int \frac{dx}{f(x)} = ln(f(x)) + C[/tex]
     
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