Splitting the integral in two ranges, each including only one of the problematic points is wise. We can easily show that both of these integrals are finite as follows:
Consider the tail of the integral extending to infinity. We know that $\arctan(x)<\pi/2$ and thus we immediately conclude that
$$\int_{A}^{\infty}\frac{e^{\arctan x}-1}{x\sqrt{x}}dx<\int_{A}^{\infty}\frac{e^{\pi/2}-1}{x\sqrt{x}}dx=2\frac{e^{\pi/2}-1}{A^{1/2}}<\infty$$
For the first part finding a suitable bound is a bit less intuitive, but it's not that hard to tell that we need to put an upper bound on the function $f(x)=(e^x-1)/x$ which is constant in the vicinity of the origin. We take the derivative of $f$
$$f'(x)=\frac{(x-1)e^x+1}{x^2}\equiv\frac{g(x)}{x^2}$$
We examine the sign of $g$. The derivative of this function is $g'(x)=xe^x$ for $x>0$ and therefore we conclude that $g$ is increasing and thus
$$g(x)> g(0)\Rightarrow f'(x)> 0 ~~\forall~~x>0 $$
We finally have concluded that $f$ is itself and increasing function, and thus we have established that an upper bound for it is given by it's value at the rightmost boundary. Since $f(x)<f(A)$ and $\arctan x<x$ we get the estimate
$$\int_{0}^{A}\frac{e^{\arctan x}-1}{x\sqrt{x}}dx<\int_{0}^{A}\frac{e^x-1}{x\sqrt{x}}dx<\frac{e^A-1}{A}\int_{0}^A\frac{dx}{\sqrt{x}}=2\frac{e^A-1}{\sqrt{A}}<\infty$$
and the integral converges to a finite value.
