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Math Help - Du Smaller than Dx Integration Problem

  1. #1
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    Du Smaller than Dx Integration Problem

    Du Smaller than Dx Problem

    \int \dfrac{2x}{x + 1}dx

    u = (x + 1)

    du = 1

    Reasoning Behind the Problem (according to the book):

    \int \dfrac{2x}{x + 1}dx = 2 \int \dfrac{x}{x + 1}dx

     = 2 \int (\dfrac{x + 1}{x + 1} - \dfrac{1}{x + 1})dx

     = 2 \int (1 - \dfrac{1}{x + 1})dx

    = 2 \int dx - 2 \int \dfrac{dx}{x + 1}

    \int \dfrac{2x}{x + 1}dx = 2x - 2 \ln(x + 1) + C Answer

    But have no clue what's going on (in any of it). Any hints? Of course I understand what to do if du = dx at the start. In that case, you do nothing. However, if du is bigger than dx than you multiply the dx by a constant, and then balance it with the reciprocal (of the constant) in the answer.
    Last edited by Jason76; January 9th 2013 at 07:30 PM.
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  2. #2
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    Re: Du Smaller than Dx Integration Problem

    You started to make a u-substitution, but you should have du=dx instead of du=1. If you solve for x in terms of u, you get x=u-1. So you have everything you need to make the substitution:

    \int \dfrac{2x}{x + 1}\, dx=\int \dfrac{2(u-1)}{u}\, du

    It's a little easier to see how to proceed now.

    The book doesn't make the substitution. It just uses x+1 "as is" instead of u. But the steps are the same.

    Hope this helps,
    Hollywood
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