Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Part 1 of the question

Hey again,

I hope that I was able to solve my problem. This is my solution with the example of Part 1 of the question. Is that correct or did I forget anything?

$$ \int \sqrt{(\sum\limits_{k=0}^{n}c_k x^k)^2} dx $$

$z_1,z_2,...,z_q$ are the real roots of $\sum\limits_{k=0}^{n}c_k x^k$

sgn(x) is the signum function

$$= \frac{x \sqrt{(\sum\limits_{k=0}^{n}c_k x^k)^2} (\sum\limits_{k=0}^{n}c_k x^k (\prod\limits_{m=0, m\neq k}^{n}m+1))}{(\prod\limits_{k=0}^{n}k+1) (\sum\limits_{k=0}^{n}c_k x^k)} + (\sum\limits_{p=1}^{q}sgn(x-z_p) \lim_{x \to z_p^-}(\frac{x \sqrt{(\sum\limits_{k=0}^{n}c_k x^k)^2} (\sum\limits_{k=0}^{n}c_k x^k (\prod\limits_{m=0, m\neq k}^{n}m+1))}{(\prod\limits_{k=0}^{n}k+1) (\sum\limits_{k=0}^{n}c_k x^k)}))$$

Example: $n=2, c_0=-1, c_1=0, c_2=1$

$z_1 = -1, z_2 = 1$

$$\int \sqrt{(\sum\limits_{k=0}^{2}c_k x^k)^2} = \sqrt{(x^2-1)^2}dx$$

$$=\frac{\sqrt{(x^2-1)^2} (x^3-3x)}{3*(x^2-1)} + \frac{2}{3} (sgn(x+1)+sgn(x-1))$$


Edit: Nobody?

share|cite|improve this question
Please, please, please take the advice I gave you yesterday, and go to those links I listed, and learn how to format mathematics for this site. Honestly, you'll be glad you did. – Gerry Myerson Nov 8 '12 at 2:33
It's latex now. – Gunnar Nov 8 '12 at 3:00
Good! Now I'd suggest removing all those asterisks --- $ab$ is better than $a*b$. – Gerry Myerson Nov 8 '12 at 3:06
Also done. no * anymore. – Gunnar Nov 8 '12 at 3:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.