Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$ Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$
$\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx=\lim\limits_{t\to\infty}\int_1^{t} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$
Partial integration can't solve the integral $\int \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$.
What substitution (or other methods) would you suggest?
 A: For $x \in [1,\infty)$, you have $$0 \le \frac{1}{x\sqrt[3]{x^2+1}} \le \frac{1}{x^\frac{5}{3}}$$ and $\int_1^\infty \frac{dx}{x^\frac{5}{3}}$ is convergent.
A: $$0\leq \int_{1}^{+\infty}\frac{dx}{x\sqrt[3]{x^2+1}} = \int_{0}^{+\infty}\frac{du}{\sqrt[3]{e^{2u}+1}}\leq \int_{0}^{+\infty}e^{-2u/3}\,du=\frac{3}{2}.$$
A: One can actually evaluate this integral "exactly" using a substitution and then partial fractions. Let $x^2+1=u^3$ so that the radical part becomes just $u.$ Also $2x\ dx = 3u^2\ du,$ and since the original integral doesn't have $2x\ dx$ in its numerator we can put it there by multiplication of numerator and denominator by $2x$ and then there will then be a $2x^2$ in the denominator, and from our substitution, $x^2=u^3-1.$ Finally when $x$ goes from $1$ to $\infty,$ our $u$ will range from $a=2^{(1/3)}$ to $\infty.$ So the starting integral has become
$$\int_a^\infty\frac{3u^2\ du}{2(u^3-1)u}$$
Now since the denominator factors into $2(u-1)(u^2+u+1)u,$ partial fractions can be applied to the $u$ integral. An involved task, but could be done.
Edit: I should have noticed that now one can cancel a $u$ from numerator and denominator, so the partial fractions to be done is simply $(3/2)[u]/[u^3-1].$
A: The integral $$\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$$ converges. 
Indeed $$\sum_{n=1}^\infty {\frac{1}{n\sqrt[3]{n^2+1}}} $$ converges. Indeed  $$ {\frac{1}{n\sqrt[3]{n^2+1}}}\sim_{\infty} {\frac{1}{n\sqrt[3]{n^2}}} $$   that converges
