Answer
The integral $\int_{1}^{\infty} \dfrac{1}{x^4+e^x} \ dx$ converges.
Work Step by Step
We are given the function
$f(x)=\int_{1}^{\infty} \dfrac{1}{x^4+e^x} \ dx$
Since, $x^4+e^x \geq x^4$
This yields:
$\dfrac{1}{x^4+e^x} \leq \dfrac{1}{x^4} $
Consider the integral
$\int_{1}^{\infty} \dfrac{1}{ x^4} dx=\lim\limits_{R \to \infty}\int_1^{R} x^{-4} dx\\=\lim\limits_{R \to \infty}[\dfrac{-1}{3x^3}]_1^R \\=\dfrac{1}{3}$
Thus, the integral $\int_{1}^{\infty} \dfrac{1}{x^4} dx$ converges to $\dfrac{1}{3}$. Therefore, by the comparison test, the integral $\int_{1}^{\infty} \dfrac{1}{x^4+e^x} \ dx$ converges as well.
