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Math Help - Integral - partial fractions?

  1. #1
    Junior Member symstar's Avatar
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    Integral - partial fractions?

    \int \frac{dx}{4x^{2/3}-4x^{1/3}-3}

    u=x^{1/3} \text{   }du=\tfrac{1}{3}x^{-2/3} \text{   }x=u^3

    =\int\frac{x^{-2/3}*3du}{(2u+1)(2u-3)}
    =3\int\frac{du}{u^2(2u+1)(2u-3)}

    Am I going about this the right way? I ask because when I use partial fractions, I am getting 2 different values for A.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by symstar View Post
    \int \frac{dx}{4x^{2/3}-4x^{1/3}-3}

    u=x^{1/3} \text{   }du=\tfrac{1}{3}x^{-2/3} \text{   }x=u^3

    =\int\frac{x^{-2/3}*3du}{(2u+1)(2u-3)}
    =3\int\frac{du}{u^2(2u+1)(2u-3)}

    Am I going about this the right way? I ask because when I use partial fractions, I am getting 2 different values for A.
    you are ok so far

    so you must be making a mistake with the partial fractions part. what did you get?
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  3. #3
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    Krizalid's Avatar
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    Quote Originally Posted by symstar View Post

    \int\frac{du}{u^2(2u+1)(2u-3)}
    Put u=\frac1z and your integral becomes -\int{\frac{z^{2}}{(2+z)( 2-3z)}\,dz}, hence

    \begin{aligned}<br />
   -\frac{z^{2}}{(2+z)(2-3z)}&=\frac{(2+z)(2-z)-4}{(2+z)(2-3z)} \\ <br />
 & =\frac{2-z}{2-3z}-\frac{4}{(2+z)(2-3z)}=\frac{2-z}{2-3z}-\frac{3(2+z)+(2-3z)}{2(2+z)(2-3z)} \\ <br />
 & =\frac{2-z}{2-3z}-\frac{1}{2}\left( \frac{3}{2-3z}+\frac{1}{2+z} \right). \\ <br />
\end{aligned}

    Things should be easy from there.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Put u=\frac1z and your integral becomes -\int{\frac{z^{2}}{(2+z)( 2-3z)}\,dz}, hence

    \begin{aligned}<br />
   -\frac{z^{2}}{(2+z)(2-3z)}&=\frac{(2+z)(2-z)-4}{(2+z)(2-3z)} \\ <br />
 & =\frac{2-z}{2-3z}-\frac{4}{(2+z)(2-3z)}=\frac{2-z}{2-3z}-\frac{3(2+z)+(2-3z)}{2(2+z)(2-3z)} \\ <br />
 & =\frac{2-z}{2-3z}-\frac{1}{2}\left( \frac{3}{2-3z}+\frac{1}{2+z} \right). \\ <br />
\end{aligned}

    Things should be easy from there.
    i know what you're thinking, symstar, "how does he do it?!"
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  5. #5
    Junior Member symstar's Avatar
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    So working with the partial fractions:

    \frac{1}{u^2(2u+1)(2u-3)}=\frac{A}{u^2}+\frac{B}{(2u+1)}+\frac{C}{(2u-3}
    1=A(4u^2-4u-3)+B(2u^3-3u^2)+C(2u^3+u^2)
    2B+2C=0
    4A-3B+C=0
    -4A=0
    -3A=1

    You can see the 2 values for A at the bottom. What should I do?
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  6. #6
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    Hello, symstar!

    I don't agree . . .


    \int \frac{dx}{4x^{\frac{2}{3}}-4x^{\frac{1}{3}}-3}

    Let: u \:=\:x^{\frac{1}{3}}\quad\Rightarrow\quad x \:=\:u^3\quad\Rightarrow\quad dx \:=\:3u^2\,du

    Substitute: . \int \frac{3u^2\,du}{4u^2 - 4u - 3} \;=\;\frac{3}{4}\int \frac{u^2\,du}{u^2-u-\frac{3}{4}} \;=\;\frac{3}{4}\int\left[1 + \frac{u+\frac{3}{4}}{u^2-u-\frac{3}{4}}\right]\,du

    . . = \;\frac{3}{4}\int\left[1 + \frac{4u+3}{4u^2-4u-3}\right]\,du \;=\;\frac{3}{4}\int\left[1 + \frac{4u+3}{(2u-3)(2u+1)}\right]\,du


    Partial Fractions: . \frac{3}{4}\int\left[1 + \frac{\frac{9}{4}}{2u-3} - \frac{\frac{1}{4}}{2u+1}\right]\,du \quad\hdots\text{ etc.}

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  7. #7
    Junior Member symstar's Avatar
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    \frac{3}{4}\int \frac{u^2\,du}{u^2-u-\frac{3}{4}} \;=\;\frac{3}{4}\int\left[1 + \frac{u+\frac{3}{4}}{u^2-u-\frac{3}{4}}\right]\,du

    I'm not quite sure what exactly happened in this step, could you explain please?
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