# Integral

• Jan 1st 2008, 04:45 PM
polymerase
Integral
$\int \frac{1}{x^7-x}\;dx$

Thanks
• Jan 1st 2008, 05:12 PM
galactus
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.

$x^{7}-x = x(x+1)(x-1)(x^{2}+x+1)(x^{2}-x+1)$
• Jan 2nd 2008, 05:52 AM
polymerase
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
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...

$\int \frac{1}{x(x^3-1)(x^3+1)}=\frac{1}{2}\int \frac{(x^3+1)-(x^3-1)}{x(x^3-1)(x^3+1)}=\frac{1}{2}\int\frac{1}{x(x^3-1)}-\frac{1}{x(x^3+1)}$
• Jan 2nd 2008, 05:58 AM
PaulRS
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: $\int{\frac{du}{6\cdot{u}\cdot{(u+1)}}}=\frac{1}{6} \cdot{\int{\frac{du}{u\cdot{(u+1)}}}}$
• Jan 2nd 2008, 06:06 AM
polymerase
Quote:

Originally Posted by PaulRS
Multiply and divide by $6\cdot{x^5}$

That was the critical step i need to know!!
Thanks
• Jan 2nd 2008, 06:09 AM
Krizalid
Quote:

Originally Posted by polymerase
$\int \frac{1}{x^7-x}\;dx$

You can also avoid substitutions

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

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

The rest follows.
• Jan 2nd 2008, 06:22 AM
Soroban
Hello, PaulRS!

That's brilliant! . . . Thank you!

• Jan 2nd 2008, 06:50 AM
galactus
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?.