## Calculus (3rd Edition)

The integral $\int_2^\infty \dfrac{dx}{x^3-4}$ converges as well by the comparison test.
Since, $x \geq2, x^3 \geq 8$ and $2x^3-8 \geq x^3$ and $x^3-4 \geq \dfrac{x^3}{2}$ So, for $x \geq 2$, it follows that $\dfrac{1}{x^3-4}\leq \dfrac{2}{x^3}$ We know that $\int_{1}^{\infty} \dfrac{dx}{x^3}$ converges by the p-test $3 \gt 1$. This means that for $\int_{1}^{\infty} \dfrac{dx}{x^3}$, the limit exists. Let us say it is $L \lt \infty$ So, $\int_{1}^{\infty} \dfrac{2}{x^{3}} dx =2 \int_{1}^{\infty} \dfrac{dx}{x^3}=2 L$ So, the integral $\int_{1}^{\infty} \dfrac{2dx}{x^3}$ converges. Hence, the integral $\int_2^\infty \dfrac{dx}{x^3-4}$ converges as well by the comparison test.