Chapter 6 - Differential Equations - 6.1 Exercises - Page 404: 44

$y=\ln(4+e^{x})$$+C$

$\frac{dy}{dx}$=$\frac{e^{x}}{(4+e^{x})}$ $y=$$\int\frac{e^{x}}{(4+e^{x})}$$dx$ if $u=(4+e^x)$, $\frac{du}{dx}$$=$$e^{x}$, then $\frac{1}{e^{x}}$$du$$=dx$ $y=$$\int\frac{e^{x}}{u}$$\frac{1}{e^{x}}$$du$ $y=\ln(u)$$+C$ $y=\ln(4+e^{x})$$+C$