A strictly increasing example: Define $g:\mathbb R_{\geq 0}\to\mathbb R_{\geq 0}$ by $g(x)=\begin{cases} n^4x-n^5+n &\text{if }n-\frac{1}{n^3}\leq x+1\leq n\\ n+n^5-n^4x &\text{if }n\leq x+1\leq n+\frac{1}{n^3}\\ 0 &\text{otherwise} \end{cases}$ which intuitively is just a continuous function which is $0$ almost everywhere but has very narrow, tall spikes at each natural number. Note that $g(n-1)=n$, so $g$ is unbounded, yet $\int_0^\infty g(x)dx = \sum\limits_{n=1}^\infty n\frac{1}{n^3}=\frac{\pi^2}{6}$ which is finite. Define $f:\mathbb R\to\mathbb R$ by $f(x)=\begin{cases} \arctan x+\int_0^x g(y)dy &\text{if } x\geq 0\\ \arctan x-\int_0^{-x} g(y)dy &\text{if } x\leq 0\\ \end{cases} $ which is continuously differentiable, bounded and strictly increasing. Since the derivative of $\arctan x$ is bounded, the derivative of $f$ differs from $g$ by a bounded function, so must be unbounded.