Please, help me to determine the following integral:
$$\int^{2}_{0} \frac{1}{\sqrt[3]{x} (x+\sqrt{x})}dx$$
As we know, this is the II order indeterminant integral. I've tried to use comparison test, and bound above the following function by $\frac{1}{x^4/3}$ but this one by the p-test is being divergent, and our integral is convergent for sure. How to proceed?
On
Substitute
$$x=\frac1u\implies dx=-\frac{du}{u^2}\implies \int_0^2\frac{dx}{x^{1/3}\left(x+x^{1/2}\right)}=\int_\infty^{1/2}-\frac{du}{u^2}\frac{u^{1/3}}{\frac1u+\frac1{u^{1/2}}}=$$
$$=\int_2^\infty\frac{u^{1/3}du}{u+u^{3/2}}\stackrel{\color{red}*}\le\int_2^\infty\frac{du}{u^{7/6}}$$
and this last integral clearly converges since $\;\frac76>1\;$ .
$\;\color{red}*\;$ This inequality follows from
$$\frac{u^{1/3}}{u+u^{3/2}}\le\frac{u^{1/3}}{u^{3/2}}=\frac1{u^{7/6}}\;,\;\;u\in (2,\infty)$$
The function $f\colon(0,2]$ defined by $$ f(x) = \frac{1}{x^{1/3}(x+x^{1/2})} $$ is continuous, so the only possible issue is at $0^+$. But at $0^+$, we have $$ f(x) = \frac{1}{x^{4/3}+x^{5/6}} \operatorname*{\sim}_{x\to 0^+} \frac{1}{x^{5/6}} $$ so is integrable by comparison, since $\frac{5}{6} < 1$.
We used the fact that $x^{4/3}+x^{5/6} = x^{5/6}(1+x^{1/2}) = x^{5/6}(1+o(1)){\displaystyle\operatorname*{\sim}_{x\to 0^+}} x^{5/6}$ when $x\to 0^+$.