the question is does this converge or diverge and if it converges give its upper bound..

$\displaystyle \int\limits_1^\infty \frac{\sin(3x)+3}{x^3+x}dx$

what i have is by comparison

$\displaystyle x^3+x>x^3$

$\displaystyle \frac{1}{x^3+x}<\frac{1}{x^3}$

and

$\displaystyle -1\leq\sin(3x)\leq1$

$\displaystyle \frac{2}{x^3+x}\leq\frac{\sin(3x)}{x^3+x}\leq\frac {4}{x^3+x}$

so

$\displaystyle \frac{\sin(3x)}{x^3+x}<\frac{4}{x^3}$

and i know that $\displaystyle \frac{4}{x^3}$ converges

but not sure how to get the upper bound, can someone explain to me how to get the upper bound please! thank you