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Math Help - Tricky Integral

  1. #1
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    Tricky Integral

    So I have \displaystyle \int{\frac{\sqrt{x}}{x+1}\,dx}

    My initial attempts at u substitution failed because I couldn't see any clear derivatives of terms in the numerator or denominator.

    I looked at Wolfram Alpha and it says to use u = \sqrt{x} and du=\frac{1}{2\sqrt{x}}

    But then, I'm not really sure where to go from there. It says to go right to

    \displaystyle \2\int{\frac{u^2}{u^2+1}\,du}

    but I can't figure out how they got that from the u and du.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Chaobunny View Post
    So I have \displaystyle \int{\frac{\sqrt{x}}{x+1}\,dx}

    My initial attempts at u substitution failed because I couldn't see any clear derivatives of terms in the numerator or denominator.

    I looked at Wolfram Alpha and it says to use u = \sqrt{x} and du=\frac{1}{2\sqrt{x}}

    But then, I'm not really sure where to go from there. It says to go right to

    \displaystyle \2\int{\frac{u^2}{u^2+1}\,du}

    but I can't figure out how they got that from the u and du.
    one way (perhaps the easiest way):

    \displaystyle u = \sqrt x

    \displaystyle \Rightarrow u^2 = x

    \displaystyle \Rightarrow 2u ~du = dx

    So your integral becomes:

    \displaystyle \int \frac {u}{u^2 + 1} \cdot 2 u~du

    \displaystyle = 2 \int \frac {u^2}{u^2 + 1}~du

    otherwise, you'd do some voodoo like multiplying the original integral by \frac {2 \sqrt x}{2 \sqrt x}, and then make the switch from x to u

    I assume you have no trouble taking it from there.
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  3. #3
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    Ah, thank you very much! I never could quite get those u substitutions with the algebraic manipulations.
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  4. #4
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    Quote Originally Posted by Chaobunny View Post
    So I have \displaystyle \int{\frac{\sqrt{x}}{x+1}\,dx}

    My initial attempts at u substitution failed because I couldn't see any clear derivatives of terms in the numerator or denominator.

    I looked at Wolfram Alpha and it says to use u = \sqrt{x} and du=\frac{1}{2\sqrt{x}}
    It ought to have given

    \displaystyle\ u=\sqrt{x}\Rightarrow\frac{du}{dx}=\frac{1}{2\sqrt  {x}}\Rightarrow\ dx=2\sqrt{x}du=2udu

    Then

    \displaystyle\int{\frac{\sqrt{x}}{x+1}}dx=\int{\fr  ac{u}{u^2+1}2u}du=2\int{\frac{u^2}{u^2+1}}du

    But then, I'm not really sure where to go from there. It says to go right to

    \displaystyle \2\int{\frac{u^2}{u^2+1}\,du}

    but I can't figure out how they got that from the u and du.
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  5. #5
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    Oh, I forgot to add the 2 in the original post. But my main problem was with the manipulation of the du and dx, which was answered. So thank you both very much. =)
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