Integration by parts for stochastic integral

Question

The integral with respect to a continuous increasing function $\alpha:[0,T]\rightarrow \mathbb{R}$ can be defined as follows. First define $\mu_\alpha([0,t]):= \alpha(t)-\alpha(0), t\in[0,T]$. Then $\mu_\alpha$ can be extended in a unique way to a measure on $\mathcal{B}([0,T])$. Next whenever a function $h$ is $\mu_\alpha$ integrable we refer to the integral of $h$ with respect to $\mu_\alpha$ as the Lebesgue-Stieltjes integral of $h$ with respect to $\alpha$ and write $$\int_{0}^{t}h(s)d\alpha(s)=\int_{[0,t]}h(s)\mu_\alpha(ds).$$ Now suppose there are two stochastic processes $A,B:[0,T]\rightarrow \mathbb{R}$ of finite variation satisfying $A(0)=B(0)=0$. Let \begin{equation} A\bullet B(t)=\int_{0}^{t}A(s)dB(s). \end{equation} In an voluntary exercise I am asked to prove the integration by parts formula for this stochastic integral given by \begin{equation} A(t)B(t)=A\bullet B(t) + B\bullet A(t), \end{equation} where as a hint we are to prove it for $A,B$ being increasing functions at first. I only ask for the solution or a hint on how to prove it for $A,B$ increasing, since we can write $A$ and $B$ as the difference of two increasing functions as they are of finite variation. All I see is that since $B$ for example is increasing, we can define $\mu_B([0,t]):=B(t)-B(0)=B(t)$, but I end up with \begin{equation} \int_{0}^{t}A(s)dB(s)=\int_{[0,t]}A(s)\mu_B(ds), \end{equation} where I do not know how to keep going. If they were both differentiable for example, I could use $dB(s)=B^{'}(s)ds$, plug this into the integral and use the integration by parts formula and then use $dA^{'}(s)ds=dA(s)$ and get exactly what I want to show, but for simply increasing and continuous functions, I am stuck and do not really know what to do. Any help is appreciated!