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?

The reduced form you gave in the end is a good starting point. Since we are within the interval $[0,1]$ we may expand the denominator as its corresponding geometric series, followed by changing the order of summation and integration and finally by termwise integration. Overall this leads us to
$$I=\int_0^1\frac{1-x^3}{1-x^5}\mathrm dx=\int_0^1\sum_{n=0}^\infty x^{5n}(1-x^3)\mathrm dx=\sum_{n=0}^\infty\int_0^1x^{5n}-x^{5n+3}\mathrm dx\\=\sum_{n=0}^\infty\left[\frac{x^{5n+1}}{5n+1}-\frac{x^{5n+4}}{5n+4}\right]_0^1
=\sum_{n=0}^\infty\left[\frac{1}{5n+1}-\frac{1}{5n+4}\right]$$
The difficult task is to find a closed-form for this sum. Of course, one could invoke the Digamma Function as Zacky quoted, but we can also note the following due the absolute convergence of the series
$$I=\sum_{n=0}^\infty\left[\frac{1}{5n+1}-\frac{1}{5n+4}\right]=1-\frac14+\frac16-\frac19+\cdots=1+\sum_{n=1}^\infty\left[\frac{1}{5n+1}-\frac{1}{5n-1}\right]$$
The latter sum falls down quite easy by considering a well-known series expansion of the cotangent function

$$\pi\cot(\pi z)~=~\frac1z+\sum_{n=1}^{\infty}\frac{2z}{z^2-n^2} $$

Thus, we rewrite the original sum as
\begin{align}
1+\sum_{n=1}^\infty\left[\frac{1}{5n+1}-\frac{1}{5n-1}\right]&=1-\sum_{n=1}^\infty\frac2{25n^2-1}\\
&=1+\frac15\left[\sum_{n=1}^\infty\frac{2\frac15}{\left(\frac15\right)^2-n^2}\right]\\
&=1+\frac15\left[\pi\cot\left(\frac\pi5\right)-5\right]\\
&=\frac\pi5\cot\left(\frac\pi5\right)
\end{align}

$$\therefore~I=\int_0^1\frac{1+x+x^2}{1+x+x^2+x^3+x^4}\mathrm dx=\sum_{n=0}^\infty\left[\frac{1}{5n+1}-\frac{1}{5n+4}\right]=\frac\pi5\sqrt{1+\frac2{\sqrt 5}}$$

