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Math Help - Integration problem

  1. #1
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    Question Integration problem

    Hi, guys, I'm a newbie,

    Could anyone help me please with this integral:

    S ( (X^2) / sqrt (X^3 + 3) ) dx
    (x squared over square root of ( x cubed plus three )

    My steps are as follow:

    U = ( x^3 + 3) du = 3x^2 dx;
    dx = du / (3x^2) -- correct??
    what next?
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  2. #2
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    Quote Originally Posted by Pupka View Post
    Hi, guys, I'm a newbie,

    Could anyone help me please with this integral:

    S ( (X^2) / sqrt (X^3 + 3) ) dx
    (x squared over square root of ( x cubed plus three )

    My steps are as follow:

    U = ( x^3 + 3) du = 3x^2 dx;
    dx = du / (3x^2) -- correct??
    what next?
    My approach to this type of integral is to recognise that 3 x^2 is the
    derivative of x^3+3, so this suggest the integral is of the form:

    integral (x^2) / sqrt (x^3 + 3) ) dx = k sqrt(x^3 + 3)

    then we differentiate the right hand side and equate that to the integrand
    to determine k.

    (d/dx) k sqrt(x^3 + 3) = k (3x^2) (1/2) /sqrt(x^3+3)

    ......................=(3k/2) [x^2/sqrt(x^3+3)]

    So we need 3k/2=1, or k=2/3, hence:

    integral (x^2) / sqrt (x^3 + 3) ) dx = (2/3) sqrt(x^3 + 3)

    RonL
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  3. #3
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    Mar 2007
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    Wow! Your reasoning is great! Thank you very much! I never used such method before... I mean where you found 2/3 using k. I'll keep in mind. I just read a chapter in my textbook on how to do this, and based on what I've understood, I solved the problem like this:

    S ( (X^2) / sqrt (X^3 + 3) ) dx
    u = (x^3 + 3)
    dx = du / (3x^2)

    so I have

    S (x^2) / ( u )^(1/2) du / (3x^2) since my dx = du / (3x^2)

    Now x^2 cancels in both numerator and denominator, 1/3 is taken up front and I am left with

    1/3 S 1 / ( u )^1/2

    = 1/3 S ( u )^-1/2, and integrating u gives us

    1/3 * (u^1/2)/(1/2) =
    1/3 * 2 *u^1/2 =
    2/3 * sqrt (x^3 + 3).

    Correct?
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