U substitution question... please help !?

integral of x(x − 3)^ (5/2) dx

by making the substitution u = x − 3

im not sure on how to do this. any help would be appreciated!!

3 Answers

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  • Best Answer

    u = x - 3

    Derive implicitly

    du = dx - 0

    du = dx

    So, x = u + 3

    u = x - 3

    and

    du = dx

    x * (x - 3)^(5/2) * dx =>

    (u + 3) * u^(5/2) * du =>

    u^(7/2) * du + 3 * u^(5/2) * du

    Integrate by using the power rule

    (1/(1 + 7/2)) * u^(7/2 + 1) + 3 * (1/(1 + 5/2)) * u^(5/2 + 1) + C =>

    (1/(9/2)) * u^(9/2) + 3 * (1/(7/2)) * u^(7/2) + C =>

    (2/9) * u^(9/2) + (6/7) * u^(7/2) + C =>

    (2/9) * (1/7) * u^(7/2) * (7 * u^(2/2) + 3 * 9 * 1) + C =>

    (2/63) * (7u + 27) * u^(7/2) + C =>

    (2/63) * (7 * (x - 3) + 27) * (x - 3)^(7/2) + C =>

    (2/63) * (7x - 21 + 27) * (x - 3)^(7/2) + C =>

    (2/63) * (7x + 6) * (x - 3)^(7/2) + C

  • 7 years ago

    u = x − 3

    x = u + 3

    dx = du

    ∫x(x − 3)^(5/2) dx = ∫(u + 3)u^(5/2) du =

    = ∫u⋅u^(5/2) du + 3∫u^(5/2) du =

    = ∫u^(7/2) du + 3∫u^(5/2) du =............... ∫uⁿ du = (1/(n + 1)) uⁿ⁺¹ + C

    = (1/(1 + 7/2)) u^(1 + 7/2) + (3/(1 + 5/2)) u^(1 + 5/2) + C

    = (2/9)u^(9/2) + (6/7) u^(7/2) + C =

    = (2/63)(7u + 27) u^(7/2) + C =

    = (2/63)(7(x - 3) + 27) (x - 3)^(7/2) + C =

    = (2/63)(7x + 6) (x - 3)^(7/2) + C =

  • 7 years ago

    int [ x ( x - 3 )^(5 / 2 ) dx ]

    Using the tabular method I get :

    = (2x / 7 )(x - 3 )^(7 / 2 ) - (4 / 63)(x - 3)^(9 / 2 ) + C

    Source(s): my brain
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