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Homework Help: Please I just need a push

  1. Oct 9, 2005 #1
    ok i got this problem:

    [tex]\int\frac{\sqrt{x-2}} {x+1} \,dx[/tex]

    then i have to let u=(x-2)^0.5 then i solve for x in terms of u

    so i get x= u^2+2
    dx=2u du

    [tex]\int\frac{\sqrt{x-2}} {x+1} \,dx=\int\frac{u}{u^2+3}\,2u\,du[/tex]

    where when i substitute i get:

    [tex]\int\frac{2u^2} {u^2+3}\,du[/tex]. after that im completely lost.
     
  2. jcsd
  3. Oct 9, 2005 #2
    Maybe:

    Pull the 2 out, then add 0 to the top (+3 - 3). Then, can you solve: [tex]2\int\frac{u^2 + 3 - 3} {u^2+3}\,du[/tex]
     
    Last edited: Oct 9, 2005
  4. Oct 9, 2005 #3
    Try making u = x+1
     
  5. Oct 9, 2005 #4
    what if i do this
    [tex]\frac{2u^2}{u^2+3}=2-\frac6{u^2+3}[/tex] i do polynomial long division. lol. but then what??
     
  6. Oct 9, 2005 #5
    It will be easier but the book says i have to use the other one.
     
  7. Oct 9, 2005 #6

    mezarashi

    User Avatar
    Homework Helper

    Then it's cool. The first time is easily integrable. The second term has a solution as well.

    Edit: Integrating the second term will be easier with substitution [tex]u = a\tan y[/tex], and the identity [tex]\tan^2 x + 1 = \sec^2 x[/tex]
     
    Last edited: Oct 9, 2005
  8. Oct 9, 2005 #7
    Did you see my edit? I think that is really easy to solve. No polynomial division needed :smile:
     
  9. Oct 9, 2005 #8
    [tex]2\int\frac{u^2 + 3 - 3} {u^2+3}\,du[/tex]


    how is that helpful?
     
  10. Oct 9, 2005 #9
    Split that up into two parts:

    (a+b)/ c = a/c + b/c


    So,
    [tex]2\int\frac{u^2 + 3 - 3} {u^2+3} du = 2(\int\1du - \int\frac{3}{u^2+3}du)[/tex]

    Which is easy to solve, unless I made some silly mistake.
     
    Last edited: Oct 9, 2005
  11. Oct 9, 2005 #10
    I am confused with differnent methods sorry I've been trying for a few hours. I dont know what to do.
     
  12. Oct 9, 2005 #11
    Gotcha mattmns, your right that is easier.


    looking at my table of integrals:

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

    where x is your u, and a is square root of 3.
     
    Last edited: Oct 9, 2005
  13. Oct 9, 2005 #12

    so if i will get as an answer this??

    [tex]2u-6\frac{1} {\sqrt{3}} tan^-1\frac{u} {\sqrt{3}},+c[/tex]
     
  14. Oct 9, 2005 #13
    change all your U's into [tex] \sqrt{x-2} [/tex], and you can anwser your own question, if we did this right, take the derivative, do you get the question you were asked in the beginning? That is always a good check to do when your not certain.
     
  15. Oct 9, 2005 #14
    You are totally right my friend, but in the book it gives me this answer

    [tex]2\sqrt{x-2}-2\sqrt{3}\tan^{-1}\left(\sqrt{\frac{x-2}3}\right)+C[/tex]
     
  16. Oct 9, 2005 #15
    hmmm the inner part of the ( ) looks good, some how I goofed off the 2 (3)^.5 part let me see.

    nope were still fine, just multiply top and bottom by square root of three. so that you go from [tex] \frac { 6} { \sqrt 3 } [/tex]

    to

    [tex] \frac { 6 \sqrt 3} { ( \sqrt 3)^2 } [/tex]

    and you get

    [tex] 2 \sqrt {3} [/tex]
     
    Last edited: Oct 9, 2005
  17. Oct 9, 2005 #16
    LoL, no problem thanks. Thank you very much for the help.

    I see i really needed it
     
    Last edited: Oct 9, 2005
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