Answer to Question #250743 in Differential Equations for Shagor

Question

1. Intrigration of dx/1+e^2x

Answer 1

$$\int\dfrac{1}{e^{2x}+1}dx=\int\dfrac{e^{-2x}}{1+e^{-2x}}dx$$

$$u=1+e^{-2x}, du=-2e^{-2x}dx$$

$$\int\dfrac{1}{e^{2x}+1}dx=\int\dfrac{e^{-2x}}{1+e^{-2x}}dx=-\dfrac{1}{2}\int\dfrac{1}{u}du$$

$$=-\dfrac{1}{2}\ln(|u|)+C=-\dfrac{1}{2}\ln(1+e^{-2x})+C$$

$$\int\dfrac{1}{e^{2x}+1}dx=-\dfrac{1}{2}\ln(1+e^{-2x})+C$$

$$=x-\dfrac{1}{2}\ln(1+e^{2x})+C$$