Here is my problem

Let $H(x)=\Phi(\frac{Ax+b}{\sqrt{2}})$ where $A$, $b$ are some non-negative constant and $\Phi$ is the CDf of a standard normal distribution.

I want to find

$$\int x \, dH(x)$$

I thought maybe integration by parts which yields:

$$x H(x)-\int H(x) \, dx$$

And now I'm not sure how to proceed or if it's possible to take the integral of a CDf
['https://math.stackexchange.com/questions/130190', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/28789/']
\begin{eqnarray}
\lim\_{x\to\infty}\left(1+\frac{f(x)}x\right)^x
&=&\lim\_{x\to\infty}\exp \left(x\log \left(1+\frac{f(x)}x \right) \right)=\exp \left(\lim\_{x\to\infty}x\log \left(1+\frac{f(x)}x\right)\right)\\ \ \\
&=&\exp \left(\lim\_{x\to\infty}x \left(\frac{f(x)}x+o\left(\frac1{x^2}\right)\right)\right)
=\exp \left(\lim\_{x\to\infty}f(x)+o\left(\frac1{x}\right)\right)\\ \ \\
&=&\exp \left(\lim\_{x\to\infty}f(x)\right).
\end{eqnarray}

The limit exchange in the second equality is justified by the fact that the exponential is continuous and that the expression inside converges.
Since there exist $\lim\limits\_{x\to 0}f(x)=l$ we have that $f$ is bounded in the neighborhood of $\infty$. So $\lim\limits\_{x\to\infty}\frac{f(x)}{x}=0$ and we can write
$$
\lim\limits\_{x\to\infty}\left(1+\frac{f(x)}{x}\right)^x=
\lim\limits\_{x\to\infty}\left(\left(1+\frac{f(x)}{x}\right)^{\frac{x}{f(x)}}\right)^{f(x)}=
\lim\limits\_{x\to\infty}\left(\left(1+\frac{f(x)}{x}\right)^{\frac{x}{f(x)}}\right)^{\lim\limits\_{x\to\infty}f(x)}=
e^l
$$