Answer
The integral $\int_{1}^{\infty} \dfrac{dx}{x^3}$ converges to $\dfrac{1}{2}$.
Hence, the integral $\int_1^\infty \dfrac{dx}{x^3+4}$ converges as well by the comparison test.
Work Step by Step
Since, $x^3+4 \geq x^3$, it follows that $\dfrac{1}{x^3+4}\leq \dfrac{1}{x^3}$
We will compute the integral as follows:
$\int_{1}^{\infty} \dfrac{dx}{x^3}= \lim\limits_{R \to \infty} \int_{1}^R \dfrac{dx}{x^3} \\=\lim\limits_{R \to \infty} [-\dfrac{x^{-2}}{2}]_1^R\\=\dfrac{1}{2}\lim\limits_{R \to \infty}(1-\dfrac{1}{R^2})\\=\dfrac{1}{2}$
So, the integral $\int_{1}^{\infty} \dfrac{dx}{x^3}$ converges to $\dfrac{1}{2}$.
Hence, the integral $\int_1^\infty \dfrac{dx}{x^3+4}$ converges as well by the comparison test.
