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  1. MHB Master
    karush's Avatar
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    #1
    w8.5.2
    $$ \int \frac{x^4}{(4-x^2)} dx
    \implies -\int\frac{x^4}{(x^2-4)} dx
    $$
    by division
    $$\displaystyle\int\dfrac{16}{x^2-4}+x^2+4 \ dx
    \implies 16\int \frac{1}{(x+2)(x-2)}+ x^2+4 \ dx $$
    $$\frac{1}{(x+2)(x-2)}=\frac{A}{x+2}+\frac{B}{x-2}
    \displaystyle = -\dfrac{1}{4\left(x+2\right)}+\dfrac{1}{4\left(x-2\right)}
    $$
    so now we have
    $$-\frac{1}{4}\int\dfrac{1}{x+2} dx
    \ \ + \ \frac{1}{4}\int\dfrac{1}{x-2} dx
    \ \ +\int {x}^{2} \ dx
    + \int 4 \ dx $$
    just seeing if going down the right trail....
    Last edited by karush; May 25th, 2016 at 20:41.

  2. MHB Master
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    #2
    Quote Originally Posted by karush View Post
    w8.5.2
    $$ \int \frac{x^4}{(4-x^2)} dx
    \implies -\int\frac{x^4}{(x^2-4)} dx
    $$
    by division
    $$\displaystyle\int\dfrac{16}{x^2-4}+x^2+4 \ dx
    \implies 16\int \frac{1}{(x+2)(x-2)}+ x^2+4 \ dx $$
    $$\frac{1}{(x+2)(x-2)}=\frac{A}{x+2}+\frac{B}{x-2}
    \displaystyle = -\dfrac{1}{4\left(x+2\right)}+\dfrac{1}{4\left(x-2\right)}
    $$
    so now we have
    $$-\frac{1}{4}\int\dfrac{1}{x+2} dx
    \ \ + \ \frac{1}{4}\int\dfrac{1}{x-2} dx
    \ \ +\int {x}^{2} \ dx
    + \int 4 \ dx $$
    just seeing if going down the right trail....
    Yes it's fine, keep going

  3. MHB Master
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    #3 Thread Author
    $$
    \displaystyle
    16\left[
    -\frac{\ln\left({\left| x-2 \right|}\right)}{4}
    +\frac{\ln\left({\left| x+2 \right|}\right)}{4} \right]
    +\frac{{x}^{3}}{3}
    +4x
    $$
    Simplify
    $$
    \displaystyle
    4\ln\left({\left| x+2 \right|}\right)
    - 4\ln\left({\left| x-2 \right|}\right)
    +\frac{{x}^{3}}{3}
    +4x
    +C
    $$
    I hope😍

  4. MHB Master
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    #4
    As you originally negated everything in the first step, I think you need the negative of this as your answer.

  5. MHB Master
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    #5 Thread Author
    What was inside the brackets got turned around making the leading term positive

  6. Perseverance
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    #6
    Quote Originally Posted by Prove It View Post
    Yes it's fine, keep going
    No, it's not fine. The '16' is not applied to $x^2+4$ as the notation used implies. Sorry to nitpick but clean notation helps to avoid errors and improves marks. In my opinion it is an area were the OP could stand to make a substantial improvement.

  7. MHB Master
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    #7
    Quote Originally Posted by greg1313 View Post
    No, it's not fine. The '16' is not applied to $x^2+4$ as the notation used implies. Sorry to nitpick but clean notation helps to avoid errors and improves marks. In my opinion it is an area were the OP could stand to make a substantial improvement.
    I'm not sure why an eight month old thread has been bumped in the first place, I'm sure in that time the OP has already completed the problem, and probably the course it was from :P

  8. MHB Master
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    #8 Thread Author
    $\textsf{Evaluate using decomposition}$
    \begin{align}
    I_{02}&=\int \frac{x^4}{(4-x^2)} dx \\
    &=-4\int \frac{1}{x+2} \, dx +4\int\frac{1}{x-2}+\int x^2 \, dx + 4\int 1 \, dx \\
    \\
    &=-4\ln{|x+2|}+4\ln{|x-2|}+\frac{x^3}{3}+4x+C
    \end{align}

    How dis...

  9. MHB Master
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    #9
    Quote Originally Posted by karush View Post
    $\textsf{Evaluate using decomposition}$
    \begin{align}
    I_{02}&=\int \frac{x^4}{(4-x^2)} dx \\
    &=-4\int \frac{1}{x+2} \, dx +4\int\frac{1}{x-2}+\int x^2 \, dx + 4\int 1 \, dx \\
    \\
    &=-4\ln{|x+2|}+4\ln{|x-2|}+\frac{x^3}{3}+4x+C
    \end{align}

    How dis...
    Let's just end this...

    $\displaystyle \begin{align*} \int{ \frac{x^4}{4 - x^2} \, \mathrm{d}x } &= -\int{ \frac{x^4}{x^2 - 4} \, \mathrm{d}x } \\ &= -\int{ \frac{ x^4 - 4\,x^2 + 4\,x^2 }{ x^2 - 4 } \, \mathrm{d}x } \\ &= -\int{ \left( \frac{ x^4 - 4\,x^2 }{ x^2 - 4 } + \frac{ 4\,x^2 }{ x^2 - 4 } \right) \,\mathrm{d}x } \\ &= -\int{ \left[ \frac{ x^2 \, \left( x^2 - 4 \right) }{x^2 - 4} + \frac{4\,x^2}{ x^2 - 4 } \right] \, \mathrm{d}x } \\ &= - \int{ \left( x^2 + \frac{4\,x^2}{x^2 - 4} \right) \,\mathrm{d}x } \\ &= -\int{ \left( x^2 + \frac{4\,x^2 - 16 + 16}{ x^2 - 4} \right) \,\mathrm{d}x } \\ &= -\int{ \left( x^2 + \frac{4\,x^2 - 16}{x^2 - 4} + \frac{16}{x^2 - 4} \right) \,\mathrm{d}x } \\ &= -\int{ \left[ x^2 + \frac{4\,\left( x^2 - 4 \right) }{x^2 - 4} + \frac{16}{x^2 - 4} \right] \,\mathrm{d}x } \\ &= -\int{ \left( x^2 + 4 + \frac{16}{x^2 - 4} \right) \,\mathrm{d}x } \\ &= -\int{ \left[ x^2 + 4 + \frac{16}{\left( x - 2 \right) \left( x + 2 \right) } \right] \,\mathrm{d}x } \end{align*}$

    Now applying partial fractions:

    $\displaystyle \begin{align*} \frac{A}{x - 2} + \frac{B}{x + 2} &\equiv \frac{16}{\left( x - 2 \right) \left( x + 2 \right) } \\ A \,\left( x + 2 \right) + B \,\left( x - 2 \right) &\equiv 16 \end{align*}$

    Let $\displaystyle \begin{align*} x = -2 \end{align*}$ to find $\displaystyle \begin{align*} -4\,B = 16 \implies B = -4 \end{align*}$.

    Let $\displaystyle \begin{align*} x = 2 \end{align*}$ to find $\displaystyle \begin{align*} 4\,A = 16 \implies A = 4 \end{align*}$, giving

    $\displaystyle \begin{align*} -\int{ \left[ x^2 + 4 + \frac{16}{ \left( x - 2 \right) \left( x + 2 \right) } \right] \,\mathrm{d}x } &= -\int{ \left( x^2 + 4 + \frac{4}{x - 2} - \frac{4}{x + 2} \right) \,\mathrm{d}x } \\ &= - \left( \frac{x^3}{3} + 4\,x + 4\ln{ \left| x - 2 \right| } - 4\ln{ \left| x + 2 \right| } \right) + C \\ &= 4\ln{ \left| x + 2 \right| } - 4\ln{ \left| x - 2 \right| } - \frac{x^3}{3} - 4\,x + C \end{align*}$

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