Question

integraçao por partes (integral de x^3.e^2x.dx)

Answer 1

∫ x³ * e^(2x)  dx 

Integrando por partes (este é o método)

u=x³  ==>du=3x² dx
∫e^(2x) dx = ∫ dv  ==> (1/2) * e^(2x) =v

∫ x³ * e^(2x)  dx  =x³* [ (1/2) * e^(2x)] - ∫  (1/2) * e^(2x) *3x² dx

∫ x³ * e^(2x)  dx  =x³* [ (1/2) * e^(2x)] -(3/2) ∫ x² e^(2x) dx

Fazendo por partes ∫ x² e^(2x) dx:

u=x² ==>du =2x dx
∫e^(2x) dx =∫ dv ==> (1/2) *e^(2x) =v

 ∫ x² e^(2x) dx = x² * [(1/2) *e^(2x)] - ∫  (1/2) *e^(2x) *2x dx
 ∫ x² e^(2x) dx = x² * [(1/2) *e^(2x)] - ∫ x e^(2x) dx

Fazendo por partes ∫ x e^(2x) dx

u= x ==> du=dx
∫e^(2x) dx =∫ dv ==> (1/2)*e^(2x) 

 ∫ x e^(2x) dx = x*[(1/2)*e^(2x) ] - ∫ (1/2)*e^(2x)  dx

 ∫ x e^(2x) dx = x*[(1/2)*e^(2x) ] -e^(2x)  

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∫ x³ * e^(2x)  dx  =x³* [ (1/2) * e^(2x)] - (3/2) *[x² * [(1/2) *e^(2x)] -[ x*[(1/2)*e^(2x) ] -e^(2x)]  ]

Answer 2

$\boxed{\boxed{\int({x}^{3}.{e}^{2x})dx}}$

faça

$u={x}^{3} \\du=3{x}^{2}dx$

$dv={e}^{2x}\\v=\int({e}^{2x})dx\\v=\frac{1}{2}{e}^{2x}$

Integral por partes

$\boxed{\int(u)dv =u.v-\int(v)du}}$

$\int({x}^{3}.{e}^{2x})dx$

${x}^{3}.\frac{1}{2}{e}^{2x}-\int(\frac{1}{2}{e}^{2x}3{x}^{2}dx$

${x}^{3}.\frac{1}{2}{e}^{2x}-\frac{3}{2}\int({e}^{2x}{x}^{2}dx$

$\int({e}^{2x}{x}^{2})dx$

$u={x}^{2} \\du=2xdx$

$dv={e}^{2x} \\ v=\frac{1}{2}{e}^{2x}dx$

$\int({e}^{2x}{x}^{2})dx$

$={x}^{2}.\frac{1}{2}{e}^{2x}-\int(\frac{1}{2}{e}^{2x}2x)dx$

$\int(\frac{1}{2}{e}^{2x}2x)dx$

$=\frac{1}{2}x{e}^{2x} -\frac{1}{4}{e}^{2x}$

$\int({e}^{2x}{x}^{2})dx$

$={x}^{2}.\frac{1}{2}{e}^{2x}-(\frac{1}{2}x{e}^{2x} -\frac{1}{4}{e}^{2x})$

$\int({x}^{3}.{e}^{2x})dx={x}^{3}.\frac{1}{2}{e}^{2x}-\frac{3}{2}[{x}^{2}.\frac{1}{2}{e}^{2x}-(\frac{1}{2}x{e}^{2x} -\frac{1}{4}{e}^{2x})] +c$