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Math Help - Integral

  1. #1
    Senior Member polymerase's Avatar
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    Integral

    \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.

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

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

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

    \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.

    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 u=x^{6}-1, \;\ du=6x^{5}dx, \;\ \frac{du}{6}=x^{5}dx, \;\ u+1=x^{6} lead to this:
    \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.

    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...

    \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: \int{\frac{dx}{x\cdot{(x^6-1)}}}

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

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

    Thus we have: \int{\frac{du}{6\cdot{u}\cdot{(u+1)}}}=\frac{1}{6}  \cdot{\int{\frac{du}{u\cdot{(u+1)}}}}
    Last edited by PaulRS; January 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 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
    \int \frac{1}{x^7-x}\;dx
    You can also avoid substitutions

    \int {\frac{1}<br />
{{x^7  - x}}\,dx}  = \int {\frac{1}<br />
{{x\left( {x^6  - 1} \right)}}\,dx}  = \int {\frac{{x^6  - \left( {x^6  - 1} \right)}}<br />
{{x\left( {x^6  - 1} \right)}}\,dx}.

    So \int {\frac{1}<br />
{{x^7  - x}}\,dx}  = \frac{1}<br />
{6}\int {\frac{{\left( {x^6  - 1} \right)'}}<br />
{{\left( {x^6  - 1} \right)}}\,dx}  - \int {\frac{1}<br />
{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, \frac{x^{5}}{x^{6}}=\frac{1}{x}

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

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

    Make the subs and we get:

    \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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