Answer
The integral $\int_{0}^{5} \dfrac{d x}{x^{1/3}+x^3} \ dx$ converges.
Work Step by Step
We are given the function
$f(x)=\int_{0}^{5} \dfrac{d x}{x^{1/3}+x^3} \ dx$
Since, $0\leq x \leq 5$ and $x^{1/3}+x^3 \geq x^{1/3}$
This yields:
$\dfrac{1}{x^{1/3}+x^3} \leq \dfrac{1}{x^{1/3}}$
Consider the integral
$\int_{0}^{5} \dfrac{d x}{ x^{1/3}}= \dfrac{3}{2} [x^{2/3}]_{0}^{5}\\= \dfrac{3}{2} (5^{2/3}-0) \\=\dfrac{3}{2} (5^{2/3})$
Thus, the integral $\int_{0}^{5} \dfrac{d x}{x^{1/3}} \ dx$ converges. Therefore, by the comparison test, the integral $\int_{0}^{5} \dfrac{d x}{x^{1/3}+x^3} \ dx$ converges as well.
