1. ## Fun Integral

how do you $\displaystyle \int \frac{1}{(x+x^2)(1-x+x^2-x^3+x^4)}\,dx$

thanks!

2. Originally Posted by polymerase
how do you $\displaystyle \int \frac{1}{(x+x^2)(1-x+x^2-x^3+x^4)}\,dx$

thanks!
Just as a first thought:
$\displaystyle (x + 1)(x^4 - x^3 + x^2 - x + 1) = x^5 + 1$

-Dan

3. Expand in partial fractions to
$\displaystyle \frac{1}{x} - \frac{1}{{5\left( {x + 1} \right)}} - \frac{{\left( { - 1 + 2x - 3x^2 + 4x^3 } \right)}}{{5\left( {1 - x + x^2 - x^3 + x^4 } \right)}}$
I will add a word about my contempt for the emphasis put partial fractions.
It took less than one minute to type the expression into a CAS then less that 10 seconds for it to tell me the ‘partial fraction decomposition’.

4. $\displaystyle \frac{1}{(x+x^2)(1-x+x^2-x^3+x^4)}\ =\ \frac{1}{x(1+x^5)}\ =\ \frac{1}{x}-\frac{x^4}{1+x^5}$

Now integrate that.

5. Thank you all for replying!! I justed wanted to know if there is an another way of integrating this function....because at first i did not see that $\displaystyle (x-1)(1...)= (1+x^5)$...so is there another way ON TOP of the methods already seen here? Because in a test situation i may not have seen that but if there is another method that i can learn to do these types of questions with, then i can impoly that method instead.

6. Do you need another method? Those partial fractions are already straightforward enough to integrate. The trick for this integral, I think, is how to arrive at these partial fractions – and I believe this is down to experience. The more of such problems you do, the quicker you’ll be able to handle similar ones in future.

Anyway, the following equations can be handy:

(i) $\displaystyle 1-x^{n+1}\ =\ (1-x)(1+x+x^2+\ldots+x^n)$

(ii) $\displaystyle 1-x^{n+1}\ =\ (1+x)(1-x+x^2-\ldots-x^n)$ if n is odd

(iii) $\displaystyle 1+x^{n+1}\ =\ (1+x)(1-x+x^2-\ldots+x^n)$ if n is even