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?

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
