So I have to check whether this:
$$\int_0^{+\infty}\frac{2x^{5/A}+\arctan\big(1+\frac{x^C}{2}\big)-e^{-Ax}}{3x+(2x)^d+\sin(3x)}dx$$
integral converges, where $d = 4, A = 3, C = 8$.
All of these trigonometric functions make me lose everything half way and make me stuck in certain period of times and got me to the point where I am too confused.
It seems that the integrand near $x=0$ behaves like $|x|^{-1}$, thus it doesn't converge. To see this, for $x\rightarrow 0$, the numerator goes to $\operatorname{arctg}(1)-1=\pi/4-1$. In the denominator, $3x$ is dominant with respect to $(2x)^4$, and $\sin(3x)\sim 3x$ for $x\rightarrow 0$ (Taylor expansion up to the first order). This means that the integrand goes to $0$ like $(\pi/4-1)/(6x)$, that is not integrable near $x=0$ from the right (nor from the left).
Towards $+\infty$, things seem to behave better, but it's useless since the integral is not convergent near $0$.