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