Simple Integral 1

Consider the following:

$\begin{aligned} y & = \frac 1{1+e^\frac 1x} \\ y'(x) & = \frac \partial{\partial x} \left(\frac 1{1+e^\frac 1x}\right) = \frac {e^\frac 1x}{x^2 \left(1+e^\frac 1x \right)^2} \\ y'(-x) & = \frac {e^{-\frac 1x}}{x^2 \left(1+e^{-\frac 1x} \right)^2} = \frac {e^{\frac 2x} \cdot e^{-\frac 1x}}{e^{\frac 2x} \cdot x^2 \left(1+e^{-\frac 1x} \right)^2} = \frac {e^\frac 1x}{x^2 \left(e^\frac 1x + 1\right)^2} = y'(x) \end{aligned}$

Therefore, $y'(x)$ is an even function. Now consider:

$\begin{aligned} I & = \int_{-1}^1 \frac \partial{\partial x} \left(\frac 1{1+e^\frac 1x}\right) dx \\ & = \int_{-1}^1 \color{#3D99F6}{y'(x)} \ dx & \small \color{#3D99F6}{\text{Since }y'(x) \text{ is even.}} \\ & = \color{#3D99F6}{2} \int_{\color{#3D99F6}{0}}^1 y'(x) \ dx \\ & = 2 \int_0^1 \frac {\partial y}{\partial x} \cdot dx \\ & = 2 \int_{y(0)}^{y(1)} dy \\ & = 2 \cdot \frac 1{1+e^\frac 1x}\bigg|_0^1 \\ & = \frac 2{1+e} \end{aligned}$

$\implies A+B+C+D = 2+1+1+1 = \boxed{5}$