Note that an antiderivative for $\frac{1}{x\+{\ⅇ}^x}$ is not available in Maple, so an appeal to Theorem 4.5.1 will be essential.

${\int }_1^{\infty }\frac{1}{x\+{\ⅇ}^x} \mathit{\ⅆ}x$ = ${\∫}_1^{\mathrm{\∞}}\frac{1}{x\+{\ⅇ}^x}\ⅆx$$$

For $x\ge 1$, $f\left(x\right)\=\frac{1}{e^x}\=e^{-x}\>g\left(x\right)\=\frac{1}{x\+e^x}$ because making the value of a denominator smaller makes the value of the fraction larger.

Since ${\int }_1^{\infty }{\ⅇ}^{-x} \mathit{\ⅆ}x$ = ${\ⅇ}^{-1}$, by conclusion (1) in Theorem 4.5.1, ${\int }_1^{\infty }\frac{1}{x\+e^x} \mathit{\ⅆ}x$ must likewise converge.