Integration problem in $\lim_{n\to\infty}\sum_{i=1}^n{{\ln(n+5i)}\over n}-\ln n$ 
Compute $$\lim_{n\to\infty}\left[\frac{\ln(n+5)+\ln(n+10)+\ln(n+15)+...+\ln(n+5n)}n\right]-\ln n$$

My attempts to this question are shown as below
$\lim_{n\to\infty}\left[\dfrac{\ln(n+5)+\ln(n+10)+...+\ln(n+5n)}n\right]-\ln n$
$=\lim_{n\to\infty}\sum_{i=1}^n{1\over n}\ln\left[{n+5i}\over n\right]$
$=\lim_{n\to\infty}\sum_{i=1}^n{1\over n}\ln\left[1+{5i\over n}\right]$
$=\int_0^1\ln(1+5x)$dx
$=x\ln(1+5x)|_0^1-\int_0^1{x\over (1+5x)}dx$
Let $u=1+5x, dx={du\over 5}$
$=\ln6-{1\over 25}\int_1^6{(u-1)\over u}du$
$=\ln6-{1\over 5}+{\ln6\over 25}$
$={26\ln6\over 5}-{1\over 5}$

I'm not sure if it is right. Please tell me if something is wrong!
 A: 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.
A: You have missed something in the derived calculations. The ending result should be $\frac{6\ln6}{5} - \frac{1}{5}$
Also, the result is pretty straightforward. Recall that :
$$\int \ln x \mathrm{d}x = x ( \ln x -1)+ C$$
Note that $1+5x$ is a linear function and this implies no changes to the end result. Thus, you can yield your desired integral solution simply by letting $x:=1+5x$ and dividing by the coefficient of $x$, thus by $5% :
$$\int_0^1 \ln(1+5x)\mathrm{d}x  = \left[ \frac{(5x+1)\left(\ln(5x+1) -1\right)}{5}\right]_0^1 = \frac{6\ln6}{5}-1 $$
