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integrate: dx/ (x^7 -x)
- Thread starter dts5044
- Start date
Hello there:
We have \(\displaystyle \int\frac{1}{x^{7}-x}dx=\int\frac{1}{x(x^{6}-1)}dx\)
Rewrite as \(\displaystyle \int\frac{x^{5}}{x^{6}(x^{6}-1)}dx\)
Now, let \(\displaystyle u=x^{6}-1, \;\ u+1=x^{6}, \;\ du=6x^{5}dx, \;\ \frac{du}{6}=x^{5}dx\)
Then, make the subs:
\(\displaystyle \frac{1}{6}\int\frac{1}{u^{2}+u}du=\frac{1}{6}\left[\int\frac{1}{u}du-\int\frac{1}{u+1}\right]du\)
Now, integrate and resub.
We have \(\displaystyle \int\frac{1}{x^{7}-x}dx=\int\frac{1}{x(x^{6}-1)}dx\)
Rewrite as \(\displaystyle \int\frac{x^{5}}{x^{6}(x^{6}-1)}dx\)
Now, let \(\displaystyle u=x^{6}-1, \;\ u+1=x^{6}, \;\ du=6x^{5}dx, \;\ \frac{du}{6}=x^{5}dx\)
Then, make the subs:
\(\displaystyle \frac{1}{6}\int\frac{1}{u^{2}+u}du=\frac{1}{6}\left[\int\frac{1}{u}du-\int\frac{1}{u+1}\right]du\)
Now, integrate and resub.
dts5044 said:integrate: dx/ (x^7 - x)
I've tried almost everything i can think of and haven't gotten anywhere. Anyone see a good starting point?
Here is a starting point. [Good one? You decide. I'm glad I don't have to do the rest of it.]
1
-------- =
x^7 - x
1
----------- =
x(x^6 - 1)
1
------------------- =
x(x^3 - 1)(x^3 + 1)
1
------------------------------------------ =
x(x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1)
Now you can use partial fractions.
..............................................galactus said:Hello there:
We have \(\displaystyle \int\frac{1}{x^{7}-x}dx=\int\frac{1}{x(x^{6}-1)}dx\)
Rewrite as \(\displaystyle \int\frac{x^{5}}{x^{6}(x^{6}-1)}dx\)
Now, let \(\displaystyle u=x^{6}-1, \;\ u+1=x^{6}, \;\ du=6x^{5}dx, \;\ \frac{du}{6}=x^{5}dx\)
Then, make the subs:
\(\displaystyle \frac{1}{6}\int\frac{1}{u^{2}+u}du=\int\frac{1}{u}du-\int\frac{1}{u+1}du\)
Now, integrate and resub.
Yes, I like that. It's definitely better than mine.
Dr. Flim-Flam
Junior Member
- Joined
- Oct 10, 2007
- Messages
- 108
For what it is worth dept.
Remembr that integrals and derivatives are inverses of each other (integration is the inverse of differentation and differentation is the inverse of integratiion), so you answer can always be checked.
For this pacticularly nasty integral, the answer is (1/6)ln|(x^6-1)/x^6| + C. Taking its deriviative will bring you back to the original integral. Excellent check when one is in doubt.
Also a good exercise in algebra manipulation for those of you who might be a little rusty in that realm.
Remembr that integrals and derivatives are inverses of each other (integration is the inverse of differentation and differentation is the inverse of integratiion), so you answer can always be checked.
For this pacticularly nasty integral, the answer is (1/6)ln|(x^6-1)/x^6| + C. Taking its deriviative will bring you back to the original integral. Excellent check when one is in doubt.
Also a good exercise in algebra manipulation for those of you who might be a little rusty in that realm.