$\displaystyle \int \frac{1}{x^7-x}\;dx$
Thanks
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.
$\displaystyle x^{7}-x = x(x+1)(x-1)(x^{2}+x+1)(x^{2}-x+1)$
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...
$\displaystyle \int \frac{1}{x(x^3-1)(x^3+1)}=\frac{1}{2}\int \frac{(x^3+1)-(x^3-1)}{x(x^3-1)(x^3+1)}=\frac{1}{2}\int\frac{1}{x(x^3-1)}-\frac{1}{x(x^3+1)}$
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: $\displaystyle \int{\frac{du}{6\cdot{u}\cdot{(u+1)}}}=\frac{1}{6} \cdot{\int{\frac{du}{u\cdot{(u+1)}}}}$
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.
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?.