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Math Help - Integration

  1. #1
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    Integration

    I think this shouldn't be difficult but I can't figure it out! >.<

    \int\displaystyle{\frac{2x}{\sqrt{x+1}}}dx

    Thank you for helping!
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  2. #2
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    Hello !
    Quote Originally Posted by Tangera View Post
    I think this shouldn't be difficult but I can't figure it out! >.<

    \int\displaystyle{\frac{2x}{\sqrt{x+1}}}dx

    Thank you for helping!
    Substitute u=\sqrt{x+1}

    \frac{du}{dx}=\frac{1}{2 \sqrt{x+1}}=\frac 1{2u}
    Thus dx=2u du



    \int \frac{2x}{\sqrt{x+1}} ~dx=\int \frac{2x}{u} \times (2u) ~du

    and x+1=u^2 \implies x=u^2-1

    Finish it ^^

    ---------------------------------------
    Otherwise, you can do it easily by parts

    u=x \implies du=dx

    v'=\frac{dx}{\sqrt{x+1}} \implies v=2 \sqrt{x+1}
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