Question

calcule a integral por partes:

∫t².$e^{5t}$ dt

Answer 1

$\int\ {t^{2} e^{5t} \, dt$

Antes de aplicar uma integral por partes é necessário aplicar uma substituição simples.

Considere: w = 5t ∴ t = w/5

                  dw = 5dt ∴ dt = dw/5

Substituindo...

= $\frac{1}{5}  \int\ {(\frac{w}{5})^{2}.e^{w} } \, dw$

= $\frac{1}{5}  \int\ {\frac{w^{2}} {25} . e^{w} } \, dw$

= $\frac{1}{125}  \int\ {{w^{2} . e^{w} } \, dw$

Agora sim devemos aplicar uma integral por partes. Cuja regra é denotada por:

$u . v - \int\ {v . du \, dt$

Escolhendo as variáveis:

u = w² → du = 2w dw

dv = e^(w)dw → v = e^(w)

Aplicando...

$\frac{1}{125} \int\ {w^{2} e^{w} \, dw$ =

$= \frac{1}{125} ( w^{2} . e^{w} - \int\ {e^{w}.(2w)} \, dw) \\= \frac{1}{125} ( w^{2} . e^{w} - 2\int\ {e^{w}.(w)} \, dw)$

Novamente seguindo o mesmo critério anterior (Algébricas = u, Exponencial = dv):

u = w → du = dw

dv = e^(w)dw → v = e^(w)

Aplicando...

$= \frac{1}{125} ( w^{2} . e^{w} - 2 [ w . e^{w} -  \int\ {e^{w}} \, dw])\\= \frac{1}{125} ( w^{2} . e^{w} - 2 [ w . e^{w} - e^{w} ])\\=  \frac{1}{125} ( w^{2} . e^{w} - 2 w . e^{w} + 2 e^{w})$

Voltando para a variável original ( t )

$=  \frac{1}{125} [(5t)^{2} . e^{5t} - 2 (5t) . e^{5t} + 2 e^{5t}]\\= \frac{1}{125} [25t^{2} . e^{5t} - 10t . e^{5t} + 2 e^{5t}]\\=  \frac{1}{125}e^{5t}  [25t^{2} - 10t + 2]$

OBS: Esse "Â" não faz parte da resposta, é apenas erro de comando do Brainly. Apenas ignore.

Answer 2

$\displaystyle\mathsf{\int t^{2}{e}^{5t} \: dt}$

$\mathsf{u={t}^{2}\to\,du=2tdt}\\\mathsf{dv={e}^{5t}dt\to~v=\dfrac{1}{5}{e}^{5t}}$

$\displaystyle\mathsf{\int\,u.dv=u.v-\int\,v.du}$

$\displaystyle\mathsf{\int\,{t}^{2}{e}^{5t}dt=\dfrac{1}{5}{t}^{2}{e}^{5t}-\dfrac{2}{5}\int\,{e}^{5t}.t\,dt}$

$\mathsf{u_{1}=t\to\,du_{1}=dt}\\\mathsf{dv_{1}={e}^{5t}dt\to\,v_{1}=\dfrac{1}{5}{e}^{5t}}$

$\displaystyle\mathsf{\int\,u_{1}\,dv_{1}=u_{1}.v_{1}-\int\,v_{1}.du_{1}}$

$\displaystyle\mathsf{\int\,t.{e}^{5t}dt=\dfrac{1}{5}t.{e}^{5t}-\dfrac{1}{5}\int\,e^{5t}dt} \\ = \mathsf{ \frac{1}{5} t.{e}^{5t} - \frac{1}{5}. \frac{1}{5} {e}^{5t} } \\ = \mathsf{ \dfrac{1}{5} {e}^{5t}(t-\dfrac{1}{5}) + k }$

$\displaystyle\mathsf{\int\,{t}^{2}{e}^{5t}dt} \\= \mathsf{\dfrac{1}{5}{t}^{2}{e}^{5t}-\dfrac{2}{5}\left[\dfrac{1}{5}e^{5t}(t-\dfrac{1}{5})\right]+k}$