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Math Help - help with another integral

  1. #1
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    help with another integral

    I just need help seeing part of how this is solved. These are my notes from class. The parts I do not understand are in red.

    \int(4x+3)\sqrt[3]{2x-5}

    u=2x-5
    du=2dx
    1/2du=dx
    2x=u+5
    2u+10=4x
    2u+13=4x+3

    I do not understand how we we concluded that if 2x=u+5 that 2u+10=4x and 2u+13 = 4x+3.



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  2. #2
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    Quote Originally Posted by gammaman View Post
    I just need help seeing part of how this is solved. These are my notes from class. The parts I do not understand are in red.

    \int(4x+3)\sqrt[3]{2x-5}

    u=2x-5
    du=2dx
    1/2du=dx
    2x=u+5
    2u+10=4x
    2u+13=4x+3

    I do not understand how we we concluded that if 2x=u+5 that 2u+10=4x and 2u+13 = 4x+3.



    2x=u+5 multiply both sides of the equation by 2

    4x=2u+10 add 3 to both sides

    4x+3=2u+13 yay
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