# Math Help - Upper bounds

1. ## Upper bounds

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

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

what i have is by comparison
$x^3+x>x^3$

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

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

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

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

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

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

2. What you actually found, it's the upper bound.

What's this? An improper integral? A series? Also, you didn't say for what $x$ you're working on.

3. Sorry it is an improper integral, but were currently in the series section.

4. Okay, then you used $\frac{\sin (3x)+3}{x^{3}+x}\le \frac{4}{x^{3}+x}<\frac{4}{x^{3}},$ and that was enough to show its convergence.