Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 513: 52

$y=16x-4e^{2x}+\frac1 4 e^{4x}+C$

$\frac{dy}{dx}=(4-e^{2x})^2$ $dy=(4-e^{2x})^2)dx$ $\int dy=\int (4-e^{2x})^2)dx$ $y+C=\int (16-8e^{2x}+e^{4x})dx$ $u=2x, \frac 1 2 du=dx, v=4x, \frac 1 4 dv=dx$ $y+C= \int 16dx -8\int e^u \frac 1 2 du+ \int e^v \frac 1 4 dv$ $y+C=16x-4e^u+\frac 1 4 e^v+C$ $y=16x-4e^{2x}+\frac1 4 e^{4x}+C$