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    integrate using substitution
    x(2x+1)^(1/2) dx u=2x +1

    = x- u^(1/2) du/2
    = (u-1)/2 u- u^(1/2) du/2
    = 1/4 [(u-1) - U^(1/2) du
    =1/4 [ -u^(3/2) = u^(1/2)
    =1/4 [-2/5u^(5/2) + 2/3u^(3/2)]
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    You have not shown us what you have tried
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    (Original post by vjosa)
    integrate using substitution
    x(2x+1)^(1/2) dx u=2x+1

    2(2x-1)^(1/2) dx u=2x-1

    3x^2 (x^3 - 2)^4 dx u=x^3 - 2

    12xe^(3x^2) dx u=3x^2

    final answers in terms of x
    We're not here to do your homework. Please post your working.
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    (Original post by vjosa)
    integrate using substitution
    x(2x+1)^(1/2) dx u=2x +1

    = x* u^(1/2) du/2
    = (u-1)/2 u* u^(1/2) du/2
    = 1/4 [(u-1)*[ u^(1/2)] du
    =1/4 [ -u^(3/2) = u^(1/2)
    =1/4 [-2/5u^(5/2) + 2/3u^(3/2)]
    You were doing fine till you got to:=1/4 [ -u^(3/2) = u^(1/2)

    It should be 1/4*Integral of [ u^(3/2)- u^(1/2)] //You add the powers//
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    (Original post by f1mad)
    You were doing fine till you got to:=1/4 [ -u^(3/2) = u^(1/2)

    It should be 1/4*Integral of [ u^(3/2)- u^(1/2)] //You add the powers//

    thanks
 
 
 
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