Question

Calcular as integrais usando o método de substituição? ​

Answer 1

Resposta:

1.

∫ (2x²+2x-3)¹⁰ *(2x+1)  dx

u=2x²+2x-3  du=4x+2=2*(2x+1)  dx

∫ (u)¹⁰ *(2x+1)  du/(2*(2x+1)

(1/2)* ∫ (u)¹⁰ du

=(1/2) u¹¹/11 + c

=(1/22)* u¹¹ + c

Como u= 2x²+2x-3

=(1/22)* (2x²+2x-3)¹¹ + c

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2.  

∫ (x³-2)^(1/7) x² dx

u=x³-2  ==>du= 3x²  dx

∫ (u)^(1/7) x² du/3x²

(1/3) * ∫ (u)^(1/7)  du

(1/3) * u^(1/7+1)/(1/7+1) + c

(7/24) * u^(8/7) + c

Como u=x³-2

(7/24) * (x³-2)^(8/7) + c

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3.

∫ x dx/ ⁵√(x²-1)          ...parece que  o índice é 5 na imagem

u=x²-1  ==>du=2x dx

∫ x (du/2x)/ ⁵√(u)  

(1/2) ∫ du/ ⁵√(u)  

(1/2) ∫  u^(-1/5)  du

(1/2) * u^(-1/5+1) / (-1/5+1)   + c

(1/2) * u^(4/5) / (4/5)   + c

(5/8) * u^(4/5)  + c

Como  u=x²-1

= (5/8) * (x²-1) ^(4/5)  + c

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4.

∫5x*√(4-3x²)  dx

u=4-3x²  ==>du=3x  dx

∫5x*√u  du/3x

(5/3)*∫ u^(1/2) du

(5/3) * u^(1/2+1)/(1+1/2)  +c

(10/3) * u^(3/2) + c

Como u= 4-3x²

=(10/3) * (4-3x²)^(3/2) + c