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Substitution process

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data

    1) antiderivative of ((t^2)+2)/((t^3)+6t+3) dt



    2) antiderivative of r(sqrt((r^2)+2))dr



    help please with these



    2. Relevant equations



    3. The attempt at a solution

    #2 let u = r^2 + 2

    du/dr = 2r

    du = 2rdr?? i dont knoww!!
     
  2. jcsd
  3. Dec 8, 2007 #2
    [tex]\int\frac{t^2 +2}{t^3 +6t +3}dt[/tex]

    [tex]\int r\sqrt{r^2+2}dr[/tex]

    correct?
     
  4. Dec 8, 2007 #3
    yes, that is correct
     
  5. Dec 8, 2007 #4

    cristo

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    For the first one, try partial fractions (probably).

    For the second, your substitution seems promising. You will have the integral [tex]\int r u^{1/2}\frac{du}{2r}=\int \frac{1}{2}u^{1/2}du[/tex]
     
    Last edited: Dec 8, 2007
  6. Dec 8, 2007 #5
    well for both, all you do is a u-substitution

    1) u-sub of your denominator

    2) u-sub of your radican, which you already did

    so let's work 2

    [tex]\int r\sqrt{r^2+2}dr[/tex]

    [tex]u=r^2 +2[/tex]
    [tex]du=2rdr \rightarrow \frac{1}{2}du=rdr[/tex]

    Rearranging your integral, do you notice that your derivative shows up in your original integrand? by that happening, you can take it out of your integral.

    [tex]\int\sqrt{r^2 +1} rdr[/tex]

    [tex]\frac{1}{2}\int\sqrt{u}du[/tex]
     
  7. Dec 8, 2007 #6
    for my answer to #2 i got (1/3)((r^2)+2)^(3/2) + c


    is that correct??
     
  8. Dec 8, 2007 #7
    correct, now your first one works out the same way, all you have to do is factor our a common term from the derivative of your u-sub.
     
  9. Dec 8, 2007 #8
    i factored out a 3 but now im lost at finishing it up
     
  10. Dec 8, 2007 #9
    i got 3 times the antiderivative of du/u dt.. i dont think thats right though
     
  11. Dec 8, 2007 #10
    [tex]u=t^3 +6t +3[/tex]
    [tex]du=3(t^2 + 1)dt \rightarrow \frac{1}{3}du=(t^2 +1)dt[/tex]

    just replace what you have with your u-sub and derivative of your u-sub.
     
    Last edited: Dec 8, 2007
  12. Dec 8, 2007 #11
    What you have to do is, divide by that 3 so that it becomes the constant for your substituted Integral.
     
    Last edited: Dec 8, 2007
  13. Dec 8, 2007 #12
    im sorry im stuck, what do i do after i find the du=3((t^2) +2)) dt
     
  14. Dec 8, 2007 #13
    Last edited: Dec 8, 2007
  15. Dec 8, 2007 #14
    so is the answer (1/3)((t^2)+2) +c ???
     
  16. Dec 8, 2007 #15
    Unfortunately, no. Have you learned about the Integral of Ln? (Natural Log)
     
  17. Dec 8, 2007 #16
    yea it becomes 1/x doesnt iit
     
  18. Dec 8, 2007 #17
    correct, so when we complete all our substitutions, we end up with:

    [tex]\frac{1}{3}\int\frac{1}{u}du[/tex]

    so now take the Integral of that and just resubstitute.
     
  19. Dec 8, 2007 #18
    i think i have it,, (1/3)ln(t^3 + 6t + 3) ?????
     
  20. Dec 8, 2007 #19
    yes, but with + C

    you can always confirm your answer by taking the derivative, gl!
     
  21. Dec 8, 2007 #20

    cristo

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    It helps if you post your working at each stage, instead of simply posting what you get as the answer. This not only helps the person checking your work, but also helps you in that you organise your thoughts into a logical progression through the problem.
     
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