Given that the value $\ln(2)$ is not the root of any polynomial with rational coefficients. For any nonnegative integer $n$, let $p_n(x)$ be the unique polynomial with integer coefficients such that $$p_n(\ln(2)) =\int_1^2 (ln(x))^n dx$$
Compute the value of the $$\sum_{n=0}^{\infty}\frac{1}{p_n(0)}$$
We first show that $$p_n(x)=2x^n-np_{n-1}(x)$$
For this, let's calculate the given integral using integrating by parts:
$$\begin{align*} p_n(\ln(2)) &= \int_1^2(\ln(2))^ndx \\ &= x(\ln(2))^n|_1^2 -\int_1^2n(\ln(x))^{n-1}dx\\&=2(ln(2))^n-n\int_1^2(\ln(x))^{n-1}dx\\&=2(\ln(2))^n-np_{n-1}(\ln(2)) \end{align*}$$
Because $\ln(2))$ cannot be the root of any polynomial with rational coefficients, it is possible to replace $\ln(2)$ with $x$ and get the recurrence
$$p_n(x)=2x^n-np_{n-1}(x)$$
Plugging $x=0$ gives $$p_n(0)=-np_{n-1}(0)$$
According to the given condition, we have $$p_0(\ln(2))=\int_1^2dx = 1$$
Therefore, $$p_0(x)=1\implies p__0(0)=1$$
Then the recurrence obtained earlier yields
$$p_n(0)=(-1)^n\cdot n!$$
Hence, the desired sum becomes
$$\sum_{n=1}^{\infty}\frac{1}{p_n(0)}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}=\boxed{\frac{1}{e}}$$
from the Taylor expansion of $e^x$ at $x=-1$.