Show that $\int_{0}^{n} \left (1-\frac{x}{n} \right ) ^n \ln(x) dx = \frac{n}{n+1} \left (\ln(n) - 1 - 1/2 -...- 1/{(n+1)} \right )$

Hint: $\sum_{k=1}^{\infty} \frac{1}{k}$ and $ \sum_{k=1}^{\infty} \frac{1}{n+k+1} $ are both infinity so you cannot write these sums separately. Instead, you should write $\lim_{N \to \infty} [-\sum_{k=1}^{N} \frac{1}{k} +\sum_{k=1}^{N} \frac{1}{n+k+1}]$. Now $[-\sum_{k=1}^{N} \frac{1}{k} +\sum_{k=1}^{N} \frac{1}{n+k+1}]$ simplifies to $\frac 1 {N+1}+\frac 1 {N+2}+..+\frac 1 {n+N+1}-(1+\frac 1 2+\frac 1 3+...+\frac 1 {n+1})$ (for $N>n+1$). Note that $\frac 1 {N+1}+\frac 1 {N+2}+..+\frac 1 {n+N+1} \to 0$ as $N \to \infty$.

Just for your curiosity

Assuming that you enjoy the gaussian hypergeometric function, there is an antiderivative $$I_n(x)=\int \left (1-\frac{x}{n} \right ) ^n \log(x)\, dx=\frac 1{n^n}\int(n-x)^n \log(x)\,dx $$ $$I_n(x)=-\frac{(n-x)^{n+1}}{n^{n+1} (n+1) (n+2)}\Big[(n-x) \, _2F_1\left(1,n+2;n+3;\frac{n-x}{n}\right)+n (n+2) \log (x) \Big]$$ $I_n(n)=0$ and then $$\int_0^n \left (1-\frac{x}{n} \right ) ^n \log(x)\, dx=\frac{n (\log (n)-\psi (n+2)-\gamma )}{n+1}=\frac{n }{n+1}\left(\log (n)-H_{n+1}\right)$$