Answer on Question#50925 - <math> - <integral calculus="">
Find ∫e2xdx\int e^{2x}dx∫e2xdx
Solution. ∫e2xdx=12∫e2xd(2x)=(t=2x)=12∫etdt=12(et+c)=12(e2x+c)\int e^{2x}dx = \frac{1}{2}\int e^{2x}d(2x) = (t = 2x) = \frac{1}{2}\int e^{t}dt = \frac{1}{2}(e^{t} + c) = \frac{1}{2}(e^{2x} + c)∫e2xdx=21∫e2xd(2x)=(t=2x)=21∫etdt=21(et+c)=21(e2x+c)
Answer. ∫e2xdx=12(e2x+c)\int e^{2x}dx = \frac{1}{2}(e^{2x} + c)∫e2xdx=21(e2x+c), where c∈Rc \in \mathbb{R}c∈R
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