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Math Help - integration by parts

  1. #1
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    integration by parts

    hi, i have to integrate the following problem by parts. i started to do it, but then i got stuck halfway through. if anyone could show me how it is done, i would be grateful.

    INT (square root of y) * (ln(y)) dy
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  2. #2
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    Quote Originally Posted by clockingly View Post
    hi, i have to integrate the following problem by parts. i started to do it, but then i got stuck halfway through. if anyone could show me how it is done, i would be grateful.

    INT (square root of y) * (ln(y)) dy
    Let u = ln(y)
    du = 1/y

    dv = sqrt(y)
    v = (2*y^(2/3))/3

    uv - int(vdu)

    [2*y^(3/2)*ln(y)]/3 - int((2*sqrt(y))/3)

    [2*y^(3/2)*ln(y)]/3 - (4*y^(3/2))/9 + C

    Which is equivalent to:

    [2*y^(3/2)*(3*ln(y) - 2)]/9 + C
    Last edited by AfterShock; February 3rd 2007 at 07:19 PM. Reason: Added the "+ C" !
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  3. #3
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    Quote Originally Posted by clockingly View Post
    hi, i have to integrate the following problem by parts. i started to do it, but then i got stuck halfway through. if anyone could show me how it is done, i would be grateful.

    INT (square root of y) * (ln(y)) dy
    You have,
    \int \sqrt{x}\ln x dx
    Let,
    u=\ln x and v'=\sqrt{x}=x^{1/2}
    Then,
    u'=1/x and v=(2/3)x^{3/2}
    Thus,
    \frac{2}{3}\ln x\cdot  x^{3/2}-\frac{2}{3}\int \frac{x^{3/2}}{x} dx
    Simplify,
    \frac{2}{3}\ln x \cdot x^{3/2}-\frac{2}{3} \int x^{1/2} dx
    You can do it from there.
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  4. #4
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    Quote Originally Posted by clockingly View Post
    hi, i have to integrate the following problem by parts. i started to do it, but then i got stuck halfway through. if anyone could show me how it is done, i would be grateful.

    INT (square root of y) * (ln(y)) dy
    By parts.
    INT.[u]dv = u*v -INT.[v]du -------(i)

    INT.[sqrt(y) *ln(y)]dy -------------(1)

    You can try two ways. I did, and I found this is the correct assumption.
    Let u = ln(y)
    Hence, du = (1/y)dy

    And so, dv = sqrt(y) dy = (y^(1/2))dy
    Hence, v = 1/(3/2) *y^(3/2) = (2/3)y^(3/2)

    INT.[sqrt(y) *ln(y)]dy -------------(1)
    Apply the assumptions, and (i),
    = ln(y)*[(2/3)y^(3/2)] -INT.[(2/3)y^(3/2)][(1/y)dy]
    = (2/3)(y^(3/2))ln(y) -(2/3)INT.[y^(1/2)]dy
    = (2/3){y^(3/2) ln(y) -(2/3)(y^(3/2))} +C
    = [(2/3)y^(3/2)]{ln(y) -2/3} +C
    = (2/3)(y)(sqrt(y))[ln(y) -2/3] +C --------------answer.
    = [ln(y) *(2/3)y^(3/2)] -
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