## Calculus (3rd Edition)

The integral $\int_{0}^{\infty} \dfrac{dx}{x^{3/2}(x+1)}$ diverges.
We are given the function $f(x)=\int_{0}^{\infty} \dfrac{dx}{x^{3/2}(x+1)}$ Since, $x^{5/2}\leq x^{3/2}$ This yields: $\dfrac{1}{x^{3/2}(x+1)} \geq \dfrac{1}{2x^{3/2}}$ Consider the integral $\int_{0}^{1} \dfrac{dx}{2x^{3/2}}=\dfrac{1}{2} [\dfrac{-2}{x^{1/2}}]_0^1 \\=\infty$ Thus, the integral $\int_{0}^{1} \dfrac{dx}{2x^{3/2}}$ diverges. Therefore, by the comparison test, the integral $\int_{0}^{\infty} \dfrac{dx}{x^{3/2}(x+1)}$ diverges as well.