Question

folosind metoda integrarii prin parti calculati: ∫(x-1)e^x dx;x∈R

Answer 1

$\int(x - 1) {e}^{x} \: dx$

$f = x - 1 = > f' = (x - 1)' = x' - 1' = 1 - 0 = 1$

$g' = {e}^{x} = > g = \int {e}^{x} \: dx = {e}^{x}$

$= (x - 1) {e}^{x} - \int1 \times {e}^{x} \: dx$

$= (x - 1) {e}^{x} - \int {e}^{x} \: dx$

$= (x - 1) {e}^{x} - {e}^{x} + C$

$= {e}^{x} (x - 1 - 1) + C$

$= {e}^{x} (x - 2) + C$