Integrate $\int [x+1] dx$ I learned about integrating GIF functions if there is definite integrals of this form: $$\int_c ^d [x+1] dx$$
But need a bit help to integrate this GIF function if there is indefinite integral of: $$\int [x+1] dx$$
p.s I searched many places including Is it possible to integrate a greatest integer function? in different form but it's hard to understand.
Need a bit help on this Calc II problem if possible.
Thank you in advance :)
 A: Using Bernoulli Polynomials
Since the integral of $\{x\}-\tfrac12$ over a unit interval is $0$, we can write
$$
\int\left(\{x\}-\tfrac12\right)\mathrm{d}x=\tfrac12\{x\}^2-\tfrac12\{x\}+C\tag1
$$
This is the beginning of the Bernoulli Polynomials. In fact, $B_1(x)=x-\frac12$ and $\frac12B_2(x)=\frac12x^2-\frac12x+\frac1{12}$.
Since $\lfloor x+1\rfloor=x+1-\{x+1\}=x+1-\{x\}$, we can use $(1)$ to get
$$
\begin{align}
\int\lfloor x+1\rfloor\,\mathrm{d}x
&=\int\left(x+\tfrac12-\left(\{x\}-\tfrac12\right)\right)\,\mathrm{d}x\tag{2a}\\
&=\tfrac12x^2+\tfrac12x-\tfrac12\{x\}^2+\tfrac12\{x\}+C\tag{2b}\\[3pt]
&=\tfrac12x^2+\tfrac12x-\tfrac12\left(x-\lfloor x\rfloor\right)^2+\tfrac12\left(x-\lfloor x\rfloor\right)+C\tag{2c}\\[3pt]
&=x\lfloor x\rfloor-\tfrac12\lfloor x\rfloor^2+x-\tfrac12\lfloor x\rfloor+C\tag{2d}
\end{align}
$$

Using a Previous Answer
Since an answer to the referenced question was
$$
\int\lfloor x\rfloor\,\mathrm{d}x=\lfloor x\rfloor x-\tfrac12\lfloor x\rfloor^2-\tfrac12[x]+C\tag3
$$
we can simply substitute $x\mapsto x+1$ to get
$$
\begin{align}
\int\lfloor x+1\rfloor\,\mathrm{d}x
&=\color{#C00}{\lfloor x+1\rfloor(x+1)}\color{#090}{-\tfrac12\lfloor x+1\rfloor^2}\color{#00F}{-\tfrac12\lfloor x+1\rfloor}+C\tag{4a}\\
&=\color{#C00}{\lfloor x\rfloor x+x+\lfloor x\rfloor+1}\color{#090}{-\tfrac12\lfloor x\rfloor^2-\lfloor x\rfloor-\tfrac12}\color{#00F}{-\tfrac12\lfloor x\rfloor-\tfrac12}+C\tag{4b}\\[3pt]
&=\lfloor x\rfloor x-\tfrac12\lfloor x\rfloor^2+x-\tfrac12\lfloor x\rfloor+C\tag{4c}
\end{align}
$$
