Question

Integração por partes.

Calcule a integral indefinida:

$\displaystyle\int x(x+1)(2x+1)e^{x(x+1)}\,dx$

Answer 1

u=x(x+1)
u=x²+x    ==>du=(2x+1)dx

∫u*(2x+1) *e^(u)  * du/(2x+1)

∫u *e^(u)  * du

v=u   ==>dv=du

e^(u) du  = dk   ==>∫e^(u) du = ∫dk ==>e^(u) =k

∫u *e^(u)  * du= u*e^(u)-∫ e^(u) du
∫u *e^(u)  * du= u*e^(u)-e^(u)  = e^(u) * (u-1) + c

Como u=x(x+1), temos então:

e^(x(x+1) * (x(x+1-1) + c   

(e^(x²+x ) )* (x²+x-1) + c     é   a   resposta

Answer 2

Olá Lukyo.

Identidades utilizadas.

$\star~~\boxed{\boxex{\mathsf{\displaystyle\int udv = uv - \int vdu}}}$

_____________________

Organizando a expressão

$\mathsf{\displaystyle\int x(x + 1)(2x + 1)e^{x^2 + x}~dx}\\\\\\\mathsf{\displaystyle\int (x^2 + x)(2x + 1)e^{x^2 + x}~dx}$

Faça

$\mathsf{u = x^2 + x}\\\\\mathsf{du = 2x + 1 dx}$

Substituindo

$\mathsf{\displaystyle\int ue^{u} du}$

Utilizando a identidade, faca as seguintes substituições.

$\mathsf{s = u~~~~ v = e^u}\\\\\mathsf{ds=du~~~~ dv = e^u}$

Substituindo na identidade, temos.

$\mathsf{\displaystyle\int ue^u = ue^u -\int e^udx}\\\\\\\mathsf{\displaystyle\int ue^u = ue^u-e^u}\\\\\\\mathsf{\displaystyle\int ue^u=e^{u}(u - 1)}\\\\\\\mathsf{\displaystyle\int x(x+ 1)(2x+1)e^{x^2+x}dx= e^{x^2+x}(x^2 + x-1)+C}}$

Duvidas ? Comente.