Using the substitution $u = x^2e^{-4x} + 3$, find $$\int\frac{x(1-2x)}{x^2+3e^{4x}} dx$$
2026-04-03 05:46:37.1775195197
How to integrate with given substitution?
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Observe that $$I\stackrel{\text{def}}{=}\int\frac{x(1-2x)}{x^2+3e^{4x}}\:dx=\int\frac{xe^{-4x}(1-2x)}{x^2e^{-4x}+3}\:dx=\int\frac{du}{2u}=\frac{1}{2}\ln{|u|}+C$$ where $u = x^2e^{-4x}+3$.