∫ e − x d x , u = − x = = u = − x = d u = − d x = ∫ e − x d x = − ∫ e u d u = − e u d u = − e − x + C {\displaystyle {\begin{aligned}&\int {e^{-x}}dx,u=-x\\[2ex]&=&=u=-x\\[2ex]&=du=-dx\\[2ex]&=\int {e^{-x}}dx\\[2ex]&=-\int {e^{u}}du\\[2ex]&=-e^{u}du\\[2ex]&=-e^{-x}+C\\[2ex]\end{aligned}}}