Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 512: 34

$-\frac 1 2 ln|7+4e^{-x}|+C$

$\int \frac 2 {7e^x+4}$ $\int \frac 2 {7e^x+4}dx=\int \frac {2e^{-x}} {7+4e^{-x}}dx$ $u=7+4e^{-x}; -\frac {du} 4==e^{-x}dx$ $\int \frac {2(-\frac{du}4)}{u}=-\frac 1 2 \int \frac {du}u$ $-\frac 1 2 ln|u|+C$ $-\frac 1 2 ln|7+4e^{-x}|+C$