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Math Help - Integration by substitution?

  1. #1
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    Integration by substitution?

    Can anyone give me any advice on solving this integral?

    ((x^2-1))/((square root(2x-1))

    Would I use u=x^2-1 then du=2x ?
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  2. #2
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    Re: Integration by substitution?

    Let u = 2x - 1 .
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  3. #3
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    Re: Integration by substitution?

    Quote Originally Posted by homeylova223 View Post
    Can anyone give me any advice on solving this integral?

    ((x^2-1))/((square root(2x-1))

    Would I use u=x^2-1 then du=2x ?
    It's often easier to pick your sub as what's in the square root as this gives the nicer \sqrt{u} when it comes to integrating.

    \dfrac{x^2-1}{\sqrt{2x-1}} = \dfrac{x^2}{\sqrt{2x-1}} - \dfrac{1}{\sqrt{2x-1}}

    For the first term sub u = 2x-1 \Leftrightarrow du = 2 dx and also x = \dfrac{1}{2}(u+1) \Leftrightarrow x^2= \dfrac{1}{4}(u+1)^2 = \dfrac{1}{4}(u^2+2u+1)

    This gives a first term of \dfrac{1}{4} \int \dfrac{u^2+2u+1}{\sqrt{u}}\ du = \dfrac{1}{4}\left( \int \dfrac{u^2}{\sqrt{u}}\ du + \int \dfrac{2u}{\sqrt{u}}\ du + \int \dfrac{1}{\sqrt{u}}\ du\right)


    When that's integrated we get \dfrac{1}{4} \left(\dfrac{2}{5}u^{5/2} + \dfrac{4}{3}u^{3/2} + 2u^{1/2} + C'\right) = \dfrac{1}{10}u^{5/2} + \dfrac{1}{3}u^{3/2} + \dfrac{1}{2}\sqrt{u} + C_1

    where C_1 = \dfrac{1}{4}C' which is the constant of integration.



    Can you integrate the second term using the same u-sub
    Last edited by e^(i*pi); August 18th 2011 at 01:08 PM. Reason: tex
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  4. #4
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    Re: Integration by substitution?

    Hello, homeylova223!

    \int \frac{x^2-1}{\sqrt{2x-1}}\,dx

    When the radicand is linear, I let u equal the entire radical.

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

    . . x^2-1 \:=\:\left(\frac{u^2+1}{2}\right)^2 - 1 \:=\:\frac{u^4 + 2u^2 - 3}{4}

    Substitute: . \int\frac{\frac{u^4+2u^2 - 3}{4}}{u}\,u\,du \;=\;\tfrac{1}{4}\int(u^4 + 2u^2 - 3)\,du

    . . =\;\tfrac{1}{4}\left(\tfrac{1}{5}u^5 + \tfrac{2}{3}u^3 - 3u\right) + C \;=\;\tfrac{u}{60}\left(3u^4 + 10u^2 - 45) + C


    Back-substitute: . \frac{\sqrt{2x-1}}{60}\bigg[3(2x-1)^2 + 10(2x-1) - 45\bigg] + C

    . . . . . . . . . . . =\;\frac{1}{60}\sqrt{2x-1}\,(12x^2 + 8x - 52) + C

    . . . . . . . . . . . =\;\frac{1}{15}\sqrt{2x-1}\,(3x^2 + 2x-13) + C

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