Improper integral: why $\int_0^1(x^2+ x^{1/3})^{-1}\,dx$ is convergent and not $\int \frac{1}{x^2}\,dx$ ???

Question

How do I show that $\int_0^1(x^2+ x^{1/3})^{-1}\,dx$ converges? I assume you show it on $(0,1]$. Can't seem to get my head around why this would be true.

Answer 1

For positive $x$ we have $\frac{1}{x^2+x^{1/3}}\lt \frac{1}{x^{1/3}}$, and $\frac{1}{x^{1/3}}$ blows up "slowly" as $x$ approaches $0$ from the right.

More formally, we show that the improper integral $$\int_0^1 \frac{dx}{x^{1/3}}$$ converges, by calculating $$\lim_{x\to 0^+}\int_0^1\frac{dx}{x^{1/3}}.$$ The integral is equal to $\frac{3}{2}(1-\epsilon^{2/3})$, and approaches $\frac{3}{2}$ as $\epsilon$ approaches $0$ from the right.

It follows by Comparison that $\int_0^1 \frac{dx}{x^2+x^{1/3}}$ converges.

As to $\int_0^1 \frac{dx}{x^2}$, we calculate $\int_\epsilon^1 \frac{dx}{x^2}$. The integral is $\frac{1}{\epsilon}-1$, and blows up as $\epsilon$ approaches $0$ from the right.