Linear First Order ODE/y' + y = 1 over (1 + exp 2 x)

Theorem
The linear first order ODE:
 * $(1): \quad y' + y = \dfrac 1 {1 + e^{2 x} }$

has the general solution:
 * $y = e^{-x} \map \arctan {e^x} + C e^{-x}$

Proof
$(1)$ is in the form:
 * $\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where $\map P x = 1$.

Thus:

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
 * $\map {\dfrac \d {\d x} } {e^x y} = \dfrac {e^x} {1 + e^{2 x} }$

and the general solution becomes:
 * $\ds y {e^x} = \int \frac {e^x} {1 + e^{2 x} } \rd x$

The integral on the can be solved by substituting:
 * $u = e^x \implies \d u = e^x \rd x$

and so:

or:
 * $y = e^{-x} \map \arctan {e^x} + C e^{-x}$