evaluation of $\int\frac{x^5}{x^5+x+1}dx$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx$
$\bf{My\; Try::}$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx = \int\frac{\left(x^5+x+1\right)-(x+1)}{x^5+x+1}dx = x-\int\frac{x+1}{x^5+x+1}dx$
Now Let $\displaystyle I = \int\frac{x+1}{x^5+x+1}dx = \int \frac{x+1}{(x^2+x+1)\cdot (x^3-x^2+1)}$
Now I Did not understand how can i solve after that
Help Required
Thanks

A very nasty solution:
You have observed that $x^5 + x + 1 = (x^2 + x + 1)(x^3 - x^2 + 1)$. This means that you can factor $x^5 + x + 1$ into linear factors :$x^5 + x + 1 = \prod_{i=1}^5 (x-\alpha_i)$, where $\alpha_i$ can be computed. Thus, for any polynomial $p(x)$ or degree less than $5$ you can find unique constants $c_i$ such that:
$$ \frac{p(x)}{x^5 + x + 1 } = \sum_{i=1}^5 \frac{c_i}{x - \alpha_i}$$
Now, each of these terms can be integrated:
$$ \int \frac{p(x)}{x^5 + x + 1 } dx = \sum_{i=1}^5  c_i \ln(x - \alpha_i).$$
So, in principle - problem solved! But please, don't ask me to do the actual computations. Wolfram can do it, though.
