Question

integral(x^2+2x)cosxdx

Answer 1

$A=\int (x^2+2x)*cos(x) \; dx$

U = x²+2x -> dU = 2x+2 dx
dV = cos(x)dx -> V = sen(x)

integrando por partes
$A=(x^2+2x)sen(x) - \int sen(x)*(2x+2)\;dx$

resolvendo a segunda integral , fazendo por partes tbm
$B=\int sen(x)*(2x+2)\;dx$

u = (2x+2) -> du = 2dx
dv = sen(x) -> v = -cos(x)

então
$B=\int sen(x)*(2x+2)\;dx\\\\B=-(2x+2)cos(x) - \int -cos(x)*2dx\\\\B= -(2x+2)cos(x)+2\int cos(x)\;dx\\\\B=-(2x+2)cos(x)+2sen(x)$

voltando na primeira integral

$A=(x^2+2x)sen(x) -[-(2x+2)cos(x)+2sen(x)]\\\\A=(x^2+2x)sen(x) +(2x+2)cos(x) - 2sen(x)\\\\A=(x^2+2x-2)sen(x) + (2x+2)cos(x) + C\\\\ \boxed{\boxed{\int (x^2+2x)*cos(x) \; dx=(x^2+2x-2)sen(x) + 2(x+1)cos(x) + C}}$