Answer
The integral $\int_{0}^{1} \dfrac{1}{x^4+\sqrt x}\ dx$ converges.
Work Step by Step
We are given the function
$f(x)=\int_{0}^{1} \dfrac{1}{x^4+\sqrt x} \ dx$
Since, $x^4+\sqrt x \geq \sqrt x$
This yields:
$ \dfrac{1}{x^4+\sqrt x} \leq \dfrac{1}{\sqrt x} $
Consider the integral
$\int_{0}^{1} \dfrac{1}{\sqrt x} dx=\int_0^1 x^{-1/2} dx \\=[2x^{1/2}]_0^1 \\=2(1-0)\\=2$
Thus, the integral $\int_{0}^{1} \dfrac{1}{\sqrt x} dx$ converges to $2$. Therefore, by the comparison test, the integral $\int_{0}^{1} \dfrac{1}{x^4+\sqrt x}\ dx$ converges as well.