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1/(1+sqrt(x)) dx

  1. Jan 14, 2010 #1
    How do you solve these 3 integrals? :

    Integral 1 : 1/(1+sqrt(x)) dx

    Integral 2: (x^3)*(e^x^2)

    Integral 3: (x*e^x)/((x+1)^2)

    I have no idea how to solve these integrals..
     
  2. jcsd
  3. Jan 14, 2010 #2

    tiny-tim

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    Hi Alexx1! :smile:

    (have an integral: ∫ and a square-root: √ and try using the X2 tag just above the Reply box :wink:)
    For 1 and 2, use the obvious substitutions. :wink:
     
  4. Jan 14, 2010 #3

    Mentallic

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    Re: Integrals

    1) let [itex]1+\sqrt{x}=u[/itex]

    2) let [itex]x^2=u[/itex] and then apply the integration by parts.

    3) Apply integration by parts by letting [itex]u=xe^x[/itex] and [itex]v'=(x+1)^{-2}[/itex]


    If you're still stuck after this, show us what you've done and we'll help you further. Good luck!
     
  5. Jan 14, 2010 #4
    Re: Integrals

    Thx! I found the third one.
    But the first and the second one, is it like

    u = 1+sqrt(x) --> du = 1/2sqrt(x) dx --> dx = 2sqrt(x) du
    u = x^2 --> du = 2xdx --> dx = du/2


    ?

    (I learned to use t = ... --> dt = ... dx)
     
    Last edited by a moderator: Jan 14, 2010
  6. Jan 14, 2010 #5

    HallsofIvy

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    Re: Integrals

    and sqrt(x)= u- 1 so dx= 2(u- 1)du

    Yes, that's right.

     
  7. Jan 14, 2010 #6
    Re: Integrals

    For the first one I become:

    integral: 2(u-1)/u du

    = 2 (integral u/u du - integral 1/u du)

    = 2 (u - ln u)

    = 2 (sqrt(x)+1 - ln (sqrt(x)+1))

    But the correct answer is: 2 (sqrt(x) - ln (sqrt(x)+1))

    What have I done wrong?
     
  8. Jan 14, 2010 #7

    D H

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    Re: Integrals

    No, that's wrong. Alex dropped a factor of x.

    The substitution [itex]u=x^2[/itex] does lead to [itex]du=2x\,dx[/itex]. Solving for dx, [itex]dx=1/(2x)\,du[/itex], not [itex]du/2[/itex].

    Alexx1: Try using this mixed form. (Alternately, look for a [itex]2x\,dx[/itex] in the integral.) using [tex]dx=1/(2\sqrt u)\,du[/tex] will just lead to confusion.
     
  9. Jan 14, 2010 #8
    Re: Integrals

    I don't know how to find a 2x dx in the integral..
    Can you explain it to me?
     
  10. Jan 14, 2010 #9

    Mentallic

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    Re: Integrals

    What you have done wrong is that you failed to realize that the extra 2 in the answer you got can be attached to the constant of integration.
    Take the derivative of both and you'll have the same result :wink:
     
  11. Jan 14, 2010 #10

    D H

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    Re: Integrals

    The integral in question is

    [tex]\int x^3\,e^{x^2}\,dx[/tex]

    Rewrite this as

    [tex]\int x^2\,e^{x^2}\,xdx[/tex]
     
  12. Jan 14, 2010 #11
    Re: Integrals

    Thanks!
     
  13. Jan 14, 2010 #12
    Re: Integrals

    If du = 2xdx than xdx = du/2 ..

    Than you get: (1/2) * integral x^2 e^u du..

    Or am I wrong?
     
  14. Jan 14, 2010 #13

    D H

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    Re: Integrals

    Correct -- but incomplete. Why did you make the u-substitution in the exponential but not for the rest of integral?
     
  15. Jan 14, 2010 #14
    Re: Integrals


    Ow you're right.. Stupid mistake.. Thank you very much, now I've solved it!

    Can you check my other post please?

    https://www.physicsforums.com/showthread.php?t=369488
     
  16. Jan 14, 2010 #15
    Re: Integrals

    Can you also help me with this integral?

    1/(1+cos(x)+sin(x)) dx
     
  17. Jan 14, 2010 #16
    Re: Integrals

    Can you also help me with this integral?

    1/(1+cos(x)+sin(x)) dx
     
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