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Thread: Integral

  1. #1
    Senior Member polymerase's Avatar
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    Integral

    $\displaystyle \int \frac{1}{x^7-x}\;dx$

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  2. #2
    Eater of Worlds
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    There are various things tou can do. Partial fractions. Maybe just a small factoring and a sub.

    $\displaystyle \int\frac{1}{x(x^{6}-1)}dx$

    Let $\displaystyle u=x^{6}-1, \;\ du=6x^{5}dx, \;\ \frac{du}{6}=x^{5}dx, \;\ u+1=x^{6}$

    This leads to $\displaystyle \frac{1}{6}\int\frac{1}{u^{2}+u}du$

    $\displaystyle \frac{1}{6}\left[\int\frac{1}{u}du-\int\frac{1}{u+1}du\right]$

    Now, continue?.

    If you feel industrious, you could do the PFD thing.

    $\displaystyle x^{7}-x = x(x+1)(x-1)(x^{2}+x+1)(x^{2}-x+1)$
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  3. #3
    Senior Member polymerase's Avatar
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    How exactly does this: Let $\displaystyle u=x^{6}-1, \;\ du=6x^{5}dx, \;\ \frac{du}{6}=x^{5}dx, \;\ u+1=x^{6}$ lead to this:
    $\displaystyle \frac{1}{6}\int\frac{1}{u^{2}+u}du$
    thats the part i didn't understand, thats why i couldn't use subst. myself.
    The aftermath is easy....

    Quote Originally Posted by galactus View Post
    If you feel industrious, you could do the PFD thing.

    $\displaystyle x^{7}-x = x(x+1)(x-1)(x^{2}+x+1)(x^{2}-x+1)$
    After i posted this, i actually solved it, but i did this before applying PF....the above would take to long...

    $\displaystyle \int \frac{1}{x(x^3-1)(x^3+1)}=\frac{1}{2}\int \frac{(x^3+1)-(x^3-1)}{x(x^3-1)(x^3+1)}=\frac{1}{2}\int\frac{1}{x(x^3-1)}-\frac{1}{x(x^3+1)}$
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  4. #4
    Super Member PaulRS's Avatar
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    Let's see

    We want: $\displaystyle \int{\frac{dx}{x\cdot{(x^6-1)}}}$

    Multiply and divide by $\displaystyle 6\cdot{x^5}$ we get: $\displaystyle \int{\frac{6\cdot{x^5}}{6\cdot{x^6}\cdot{(x^6-1)}}}dx$

    We now let:$\displaystyle u=x^6-1\rightarrow{\frac{du}{dx}=6\cdot{x^5}}$

    Thus we have: $\displaystyle \int{\frac{du}{6\cdot{u}\cdot{(u+1)}}}=\frac{1}{6} \cdot{\int{\frac{du}{u\cdot{(u+1)}}}}$
    Last edited by PaulRS; Jan 2nd 2008 at 06:19 AM.
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  5. #5
    Senior Member polymerase's Avatar
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    Quote Originally Posted by PaulRS View Post
    Multiply and divide by $\displaystyle 6\cdot{x^5}$
    That was the critical step i need to know!!
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  6. #6
    Math Engineering Student
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    Quote Originally Posted by polymerase View Post
    $\displaystyle \int \frac{1}{x^7-x}\;dx$
    You can also avoid substitutions

    $\displaystyle \int {\frac{1}
    {{x^7 - x}}\,dx} = \int {\frac{1}
    {{x\left( {x^6 - 1} \right)}}\,dx} = \int {\frac{{x^6 - \left( {x^6 - 1} \right)}}
    {{x\left( {x^6 - 1} \right)}}\,dx}$.

    So $\displaystyle \int {\frac{1}
    {{x^7 - x}}\,dx} = \frac{1}
    {6}\int {\frac{{\left( {x^6 - 1} \right)'}}
    {{\left( {x^6 - 1} \right)}}\,dx} - \int {\frac{1}
    {x}\,dx}$.

    The rest follows.
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  7. #7
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    Hello, PaulRS!

    That's brilliant! . . . Thank you!

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  8. #8
    Eater of Worlds
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    Here's what I was getting at:

    Because, $\displaystyle \frac{x^{5}}{x^{6}}=\frac{1}{x}$

    Now, because $\displaystyle u=x^{6}-1, \;\ u+1=x^{6}, \;\ du=6x^{5}dx, \;\ \frac{du}{6}=x^{5}dx$

    Rewrite as : $\displaystyle \frac{1}{6}\int\frac{x^{5}}{x^{6}(x^{6}-1)}dx$

    Make the subs and we get:

    $\displaystyle \frac{1}{6}\int\frac{1}{(u+1)u}du=\frac{1}{6}\int\ frac{1}{u^{2}+u}du$

    See now?.
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