[br]\begin{aligned}f(x)&=\frac{xe^{2x}}{x+k}=\frac{u(x)}{v(x)}\\ [br]f^\prime(x)&=\frac{v{\rm{d}}u-u{\rm{d}v}}{v^2}\quad&=\frac{(x+k) \left[ e^{2x}(2x+1)\right]-xe^{2x}.1 }{(x+k)^2 }\\[br]\end{aligned}
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