Apologies if I’ve just missed a duplicate, couldn’t find anything.
Have tried a few things but can’t seem to figure an obvious approach for a closed-form solution. This is just the standard expansion for k successes from N trials, but we weight each term by $1/k$
The following answer gives a closed form for the coefficients of these polynomials.
If $P_n(x)=\sum_{k=1}^n\frac{\binom{n}{k}x^k(1-x)^{n-k}}{k}$, then $$P_n(x)=\sum_{k=1}^n\binom{n}{k}(-1)^{k+1}H_kx^k.$$ Where $H_k$ is the $k$th harmonic number.
\begin{align*} P_n(x)&=\sum_{k=1}^n\frac{\binom{n}{k}x^k(1-x)^{n-k}}{k}\\ &=\sum_{k=1}^n\frac{1}{k}\binom{n}{k}\sum_{r=0}^{n-k}\binom{n-k}{r}(-1)^rx^{k+r}\\ &=\sum_{k=1}^n\frac{1}{k}\binom{n}{k}\sum_{r=k}^{n}\binom{n-k}{r-k}(-1)^{r-k}x^r\\ &=\sum_{r=1}^n\sum_{k=1}^{r}\frac{1}{k}\binom{n}{k}\binom{n-k}{r-k}(-1)^{r-k}x^r\\ &=\sum_{r=1}^nx^r\sum_{k=1}^{r}\frac{1}{k}\binom{n}{k}\binom{n-k}{r-k}(-1)^{r-k}\\ \end{align*} So the coefficient of $x^r$ in $P_n(x)$ is \begin{align*} \sum_{k=1}^{r}\frac{1}{k}\binom{n}{k}\binom{n-k}{r-k}(-1)^{r-k}&=\sum_{k=1}^r\frac{(-1)^{r-k}}{k}\cdot\frac{n!}{k!(n-k)!}\cdot\frac{(n-k)!}{(r-k)!(n-r)!}\\ &=\binom{n}{r}(-1)^{r+1}\sum_{k=1}^r\frac{(-1)^{k+1}}{k}\binom{r}{k}\\ &=\binom{n}{r}(-1)^{r+1}[z^r]\left\{\frac{1}{1-z}\log\left(1+\frac{z}{1-z}\right)\right\}\\ &=\binom{n}{r}(-1)^{r+1}[z^r]\left\{\frac{1}{1-z}\log\left(\frac{1}{1-z}\right)\right\}\\ &=\binom{n}{r}(-1)^{r+1}H_r.\\ \end{align*}