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Math Help - Integration by substitution thread 2?

  1. #1
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    Integration by substitution thread 2?

    I am wondering if I did this correctly.

    Integrate

    x square root(2x+1)

    u=2x+1 x=(1/2)(u-1)

    (1/2)(u-1) square root(u)

    (1/2) (u)^(3/2)-(u)^(1/2)

    (1/2) (2/5)(u)^(5/2)-(2/3)(u)^(3/2)

    (2/10)u^(5/2)-(2/6)(u)^(3/2)+c

    Any help would be appreciated on this difficult topic.
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  2. #2
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    Re: Integration by substitution thread 2?

    Not quite. You are trying to integrate \displaystyle \int{x\sqrt{2x + 1}\,dx}.

    If you let \displaystyle u = 2x + 1 then \displaystyle x = \frac{1}{2}(u - 1) AND \displaystyle du = 2\,dx.

    So the integral becomes

    \displaystyle \begin{align*}\int{x\sqrt{2x + 1}\,dx} &= \frac{1}{2}\int{x\sqrt{2x + 1}\cdot 2\,dx} \\ &= \frac{1}{2}\int{\frac{1}{2}(u - 1)\sqrt{u}\,du} \\ &= \frac{1}{4}\int{(u - 1)u^{\frac{1}{2}}\,du} \\ &= \frac{1}{4}\int{u^{\frac{3}{2}} - u^{\frac{1}{2}}\,du} \end{align*}

    Go from here
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  3. #3
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    Re: Integration by substitution thread 2?

    Hello, homeylova223!

    \int x\sqrt{2x+1}\,dx

    Try the method I suggested in your previous post.

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

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

    . . . . =\;\tfrac{1}{2}\left(\tfrac{1}{5}u^5 - \tfrac{1}{3}u^3\right) + C \;=\; \frac{u^3}{30}(3u^2-5) + C

    Back-substitute: . \frac{(\sqrt{2x+1})^3}{30}\bigg[3(2x+1)-5\bigg] + C

    . . =\;\tfrac{1}{30}(2x+1)^{\frac{3}{2}}(6x-2) + C \;=\;\tfrac{1}{15}(2x+1)^{\frac{3}{2}}(3x-1) + C

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