Convergence of $\int_1^7 \frac{1}{\sqrt[3]{x^2-1}}\ dx$

Question

Is $$\int_1^7 \frac{1}{\sqrt[3]{x^2-1}}\ dx < \infty \ ?$$ Should I use asymptotic criterium? It's the only thing that comes to my mind, but I can't find suitable function...

Answer 1

The only are of concern is, of course, near $x=1$. Let $u=x^2-1$; then $dx = \frac{du}{2\sqrt{u+1}}$. The integral has tranformed to $$ \int_{0}^{48} \frac{u^{-\frac{1}{3}}}{2\sqrt{u+1}} du $$ The behavior of $\frac{1}{2\sqrt{u+1}}$ for $u$ near zero is $\frac{1}{2} + O(u)$ so the behavior of the integral is like that of $$ \int_{0}^{x} \frac{u^{-\frac{1}{3}}}{2} du $$ which for small positive $x$ goes like $\frac{3}{4}x^{+\frac{2}{3}}$. Since that exponent is positive, the contribution to the original interval of the region very near zero is zero, and the original integral converges.