Integration problem in $\lim_{n\to\infty}\sum_{i=1}^n{{\ln(n+5i)}\over n}-\ln n$

I'm going to assume you seek to compute $$\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{\ln(n+5k)}{n}\right)-\ln n$$ Note that by bringing $\ln n$ inside the sum, it becomes $\frac{\ln n}{n}$, so that we now want to compute $$\lim_{n\to\infty}\sum_{k=1}^n\frac{\ln(n+5k)-\ln(n)}{n}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n}\ln\left(\frac{n+5k}{n}\right)=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n}\ln\left(1+5\cdot\frac{k}{n}\right).$$ Recognizing $\Delta x=\frac{1}{n}$ and $a=0$, we can realize this last sum as a definite integral over the interval $[0,1]$ of the function $\ln(1+5x)$. That is, we need to compute $$\int_0^1\ln(1+5x)\,dx.$$ Using a $u$ substitution with $u=5x+1$, we have $$\frac{1}{5}\int_1^6\ln(u)\,du=\frac{1}{5}\left(u\ln u-u\big|_{u=1}^{u=6}\right)=\frac{6\ln 6}{5}-1.$$ This last value is the value that the original limit converges to; that is, we have $$\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{\ln(n+5k)}{n}\right)-\ln n=\frac{6\ln 6}{5}-1.$$

Mistakes Specific to your Attempt

You began to solve the integral using integration by parts; when you did this, though, your derivative of $\ln(1+5x)$ should have been $\frac{5}{1+5x}$ (using the chain rule). It appears you ended up with just $\frac{1}{1+5x}$, forgetting the factor of $5$ in the numerator.