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Math Help - integral with dx on top?

  1. #1
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    Post integral with dx on top?

    dx/(2 √x +2x)

    I tried 1/(2 √x +2x)dx
    then i know that it is not du/u so i tried finding u' but i don't think i have the right one...
    i found 6x^1/2 because the bottom is 4x^3/2
    so: -6x^1/2 6x^1/2/(2 √x +2x)dx
    then: -6x^1/2[ln(abs value 4x^3/2)+c]

    but i don't know after that.

    the answer is ln(1+ √x) +c
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by genlovesmusic09 View Post
    dx/(2 √x +2x)

    I tried 1/(2 √x +2x)dx
    then i know that it is not du/u so i tried finding u' but i don't think i have the right one...
    i found 6x^1/2 because the bottom is 4x^3/2
    so: -6x^1/2 6x^1/2/(2 √x +2x)dx
    then: -6x^1/2[ln(abs value 4x^3/2)+c]

    but i don't know after that.

    the answer is ln(1+ √x) +c
    \int\frac{\,dx}{2\sqrt{x}+2x}=\tfrac{1}{2}\int\fra  c{\,dx}{\sqrt{x}+\left(\sqrt{x}\right)^2}.

    Let {\color{red}u=\sqrt{x}}\implies\,du=\frac{\,dx}{2\  sqrt{x}}\implies 2\sqrt{x}\,du=\,dx.

    So we have \tfrac{1}{2}\int\frac{2{\color{red}\sqrt{x}}\,du}{  u+u^2}=\int\frac{u\,du}{u\left(1+u\right)}=\int\fr  ac{\,du}{1+u}

    Can you continue?
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