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

Thanks

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- Jan 1st 2008, 04:45 PMpolymeraseIntegral
$\displaystyle \int \frac{1}{x^7-x}\;dx$

Thanks - Jan 1st 2008, 05:12 PMgalactus
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.

$\displaystyle x^{7}-x = x(x+1)(x-1)(x^{2}+x+1)(x^{2}-x+1)$ - Jan 2nd 2008, 05:52 AMpolymerase
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....

After i posted this, i actually solved it, but i did this before applying PF....the above would take to long...

$\displaystyle \int \frac{1}{x(x^3-1)(x^3+1)}=\frac{1}{2}\int \frac{(x^3+1)-(x^3-1)}{x(x^3-1)(x^3+1)}=\frac{1}{2}\int\frac{1}{x(x^3-1)}-\frac{1}{x(x^3+1)}$ - Jan 2nd 2008, 05:58 AMPaulRS
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: $\displaystyle \int{\frac{du}{6\cdot{u}\cdot{(u+1)}}}=\frac{1}{6} \cdot{\int{\frac{du}{u\cdot{(u+1)}}}}$ - Jan 2nd 2008, 06:06 AMpolymerase
- Jan 2nd 2008, 06:09 AMKrizalid
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. - Jan 2nd 2008, 06:22 AMSoroban
Hello, PaulRS!

That's brilliant! . . . Thank you!

- Jan 2nd 2008, 06:50 AMgalactus
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?.