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  1. #1
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    integrate

    integrate \frac{4x}{\sqrt{2x+1}}
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  2. #2
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    \int{\frac{4x}{\sqrt{2x+1}}\,dx}.

    Let u = 2x + 1 so that x = \frac{u-1}{2} and du = 2\,dx.


    The integral becomes:

    \int{\frac{4x}{\sqrt{2x+1}}\,dx} = \int{\frac{2x}{\sqrt{2x + 1}}\,2\,dx}

     = \int{\frac{u - 1}{\sqrt{u}}\,du}

     = \int{u^{\frac{1}{2}} - u^{-\frac{1}{2}}\,du}

     = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} - \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C

     = \frac{2\sqrt{(2x + 1)^3}}{3} - 2\sqrt{2x+1} + C

     = \frac{2\sqrt{(2x + 1)^3} - 6\sqrt{2x + 1}}{3} + C

     = \frac{2(2x + 1)\sqrt{2x + 1} - 6\sqrt{2x + 1}}{3} + C

     = \frac{(4x + 2)\sqrt{2x + 1} - 6\sqrt{2x + 1}}{3} + C

     = \frac{(4x - 4)\sqrt{2x + 1}}{3} + C

     = \frac{4(x - 1)\sqrt{2x + 1}}{3} + C.
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  3. #3
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    Hello, Punch!

    Another method . . .


    Integrate: . \displaystyle{\int \frac{4x}{\sqrt{2x+1}}}\,dx

    We have: . \displaystyle{4\int\frac{x}{\sqrt{2x+1}} \,dx<br />
}


    Let: . u \:=\:\sqrt{2x+1} \quad\Rightarrow\quad x \:=\:\frac{1}{2}(u^2-1) \quad\Rightarrow\quad dx \:=\:u\,du


    Substitute: . \displaystyle{4\int\frac{\frac{u^2-1}{2}}{u}\cdot u\,du \;=\;2\int(u^2-1)\,du}

    . . . . . . . =\;2\left(\frac{u^3}{3} - u\right) + C \;=\;\frac{2}{3}u(u^2-3) + C


    Back-substitute: . \frac{2}{3}\sqrt{2x+1}\,(2x+1 - 3) + C

    . . . . . . . . . . =\; \frac{2}{3}\sqrt{2x+1}\,(2x - 2) + C

    . . . . . . . . . . =\; \frac{4}{3}\sqrt{2x+1}\,(x-1) + C


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Some advice (welcome or not):

    When given a linear expression under a radical,
    . . let u \,=\text{ the entire radical}


    Given: .an integral with \sqrt[n]{ax+b}

    \text{Let: }\:u \:=\:\sqrt[n]{ax+b} \quad\Rightarrow\quad x \:=\:\dfrac{u^n - b}{a} \quad\Rightarrow\quad dx \:=\:\frac{n}{a}u^{n-1}du

    . . and the new integral will have no radicals (usually).
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    And another solution...

    By substitution: x=(1/2)cos(2t). dx=-sin(2t), sqrt(2x+1)=sqrt(cos(2t)+1)=sqrt(2cos^2(t))=sqrt(2) *cos(t)

    Ok... maybe it's a little bit complicated...
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