Question

Integrala din (x+2018)^2 dx

Answer 1

Explicație pas cu pas:

$\int {(x + 2018)}^{2} dx$

$u = x + 2018 = > du = 1dx$

$\int {(u)}^{n} du = \frac{ {u}^{n + 1} }{n + 1} = > \int {(u)}^{2} du = \frac{ {u}^{3} }{3}$

=>

$\int {(x + 2018)}^{2} dx = \frac{ {(x + 2018)}^{3} }{3} + c$