Convergence or divergence of integral

I'm struggling with how to show that

$$\int_1^\infty \frac{x \sin x}{\sqrt{1+x^5}}dx$$ either diverges or converges.

If we call the integrand $f(x)$ then $$f(x)\leq g(x)=\frac{x}{\sqrt{1+x^5}}\forall x\in[1, \infty)$$ so I tried to show that $g(x)$ converges (and hence that $f(x)$ must converge), but with no luck. Another idea I had was to show that

$$h(x)=\frac{\sin x}{\sqrt{1+x^5}}$$ diverges, which would mean that $f(x)$ diverges too since

$$h(x)\leq f(x)\forall x\in[1, \infty)$$ but I got nowhere.

I also tried to use some partial integration and get two expressions rather than one, but that didn't help.

Can I get some hints? Please note that it's homework so no solutions, thanks!

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Note that your integrand is not positive, so you have to use absolute values. For the estimate, you have to get rid of the $1$.

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Why is it positive? $\sin x$ is between -1 and 1, and everything else is positive. – hejseb Nov 8 '13 at 11:39
Sorry, I meant to say "not positive". – Martin Argerami Nov 8 '13 at 11:43
I see, that makes more sense. I think I got it now. Thanks :) – hejseb Nov 8 '13 at 11:46
You are welcome. The suggestion was an attempt to lead you to the strategy now shown in the other answer. – Martin Argerami Nov 8 '13 at 11:49
I appreciate that. One question: if I show that the absolute value of the function converges, does that imply that it's convergent (in the 'non-absolute' sense)? Absolute convergence is discussed in later chapters for some reason, so it's a bit of a new concept... – hejseb Nov 8 '13 at 12:53

It is (absolutely) convergent by the comparison test since $\int_1^\infty|\frac{x\sin x}{\sqrt{1+x^5}}|dx\leq \int_1^\infty\frac{1}{x^{3/2}}dx$ (convergent by $p$ test since $p=3/2>1$.

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Thanks, I think I am able to develop some reasoning now why it should converge. – hejseb Nov 8 '13 at 11:47