How would I compute this integral with a ceiling function? It seems simple, but it is not, really. How would I calculate this ($\lceil a \rceil$ denotes  the ceiling function)? 
$$\int{\lceil x+2 \rceil}\ln x\,\,dx$$
First, I noticed that $\lceil x+2 \rceil=\lceil x \rceil + 2$. So I got:
$$\int({\lceil x \rceil}+2)\ln x\,\,dx=\int2\ln x\,\,dx+\int{\lceil x \rceil}\ln x\,\,dx$$
Then I got:
$$\int2\ln x\,\,dx=2\int\ln x\,\,dx=2x\ln x-2x$$
So then:
$$2x\ln x-2x+\int{\lceil x \rceil}\ln x\,\,dx$$
From this point, I tried integrating by parts. So first, I substituted:
$$u=\ln x\qquad dv=\lceil x \rceil$$$$du=\frac1x\qquad v=\int\lceil x \rceil=\,?$$
So I'm left with:
$$2x\ln x-2x+\Bigg(\ln x\int\lceil x\rceil-\int\frac1x\Bigg(\int\lceil x\rceil\Bigg)\Bigg)$$
But now I'm stuck. Am I doing correct so far? How do I even calculate the integral of $\lceil x\rceil$?

$$
\begin{align}
\int\lceil x+2\rceil\log(x)\,\mathrm{d}x
&=\int\lceil x+2\rceil\,\mathrm{d}(x\log(x)-x)\\
&=\lceil x+2\rceil(x\log(x)-x)-\int(x\log(x)-x)\,\mathrm{d}\lceil x+2\rceil\\
&=\lceil x+2\rceil(x\log(x)-x)-\sum_{k=1}^{\lceil x-1\rceil}(k\log(k)-k)+C
\end{align}
$$

$$
\begin{align}
\int\lceil x\rceil\,\mathrm{d}x
&=x\lceil x\rceil-\int x\,\mathrm{d}\lceil x\rceil\\
&=x\lceil x\rceil-\sum_{k=0}^{\lceil x-1\rceil}k+C\\
&=x\lceil x\rceil-\frac{\lceil x\rceil^2-\lceil x\rceil}2+C
\end{align}
$$
