Evaluate $\int_{0}^{1}\frac{1+x+x^2}{1+x+x^2+x^3+x^4}dx$ Evaluate $$I=\int_{0}^{1}\frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx$$
My try:
We have:
$$1+x+x^2=\frac{1-x^3}{1-x}$$
$$1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$$
So we get:
$$I=\int_{0}^{1}\frac{1-x^3}{1-x^5}dx$$
$$I=1+\int_{0}^{1}\frac{x^3(x^2-1)}{x^5-1}dx$$
Any idea from here?

Your integral $$ \int_{0}^{1}\frac{1-x^3}{1-x^5}dx$$ is similar to that of the digamma function $$-\gamma+\int_{0}^{1}\frac{1-t^s}{1-t}dt=\psi(s + 1) $$Maybe the change of variable $t=x^5$ could be helpful.
