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Integración por partes

Module by: Beatriz Nava. E-mail the authorEdited By: Evaristo Rojas Mayoral

Summary: Ejercicio resuelto por el método de integración por partes

Integración por partes

Ejemplo:

x²dx1+x = 215 (3x²-4x+8) x+1 + C Fórmula de Integración por partes:

udv = uv-v du

Desarrollo:

u= X² dv= 1 1+xdx

du= 2x v=11+xdx= dww= w-1/2dw = 2w1/2+ C = 21+x + C

w= 1+x

dw= dx

= X²2 (1+X)1/2 - ∫2(1+X)1/22xdx

=2x² (1+X)1/2 - 4 x(1+X)1/2dx

u= X dv= (1+X)1/2

du= dx v= (1+X)1/2dx= (1+X)3/23/2 + C= 23(1+X)3/2 + C

=2x² (1+x)1/2 - 4[ x (23 (1+X)3/2- 23(1+X)3/2dx]

=2x² (1+x)1/2 - 4[ x (23 (1+X)3/2- 23(1+X)3/2dx]

=2x² (1+x)1/2 - 83 x (1+X)3/2 + 83 (25(1+X)5/2 + C

=2x² (1+x)1/2 - 83 x (1+X)3/2 + 1615 (1+X)5/2 + C

=(1+x)1/2 [2x²- 83 x (1+X) + 1615 (1+ 2x + x²)] + C

=(1+x)1/2 [2x²- 83 (1+X²) + 1615 (1+ 2x + x²)] + C

= 1+X [2x² - 83 x - 83 x² + 1615 + 3215 x + 1615 x²] + C

=1+X [3015 x² - 4015x - 4015x²+ 1615 + 3215 x + 1615 x²] + C

=1+X [1615 x²- 815x + 1615] + C =1+X215 [3x² - 4x + 8] + C Resultado

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