Question

O valor da integral ∫1-0 4x^3(x^4+1)^6dx

Answer 1

Boa tarde


F(x) = ∫ 4x³ * (x⁴ + 1)⁶ dx

F(x) = 4 (x²⁸/28 + x²⁴/4 + (3x²⁰)/4 + (5x¹⁶)/4 + (5x¹²)/4 + (3x⁸)/4 + x⁴/4) + C 
F(1) = 4 * (1/28 + 1/4 + 3/4 + 5/4 + 5/4 + 3/4 + 1/4 )
F(1) = 4/28 + 1 + 12/4 + 20/4 + 20/4 + 12/4 + 4/4
F(1) = 1/7 + 1 + 3 + 5 + 5 + 3 + 1 = 1/7 + 18 = 127/7 

Answer 2

Podemos utilizar a integração por substituição.

$\boxed{u~=~x^4+1}\\\\\\\dfrac{d}{dx}\left(u\right)~=~\dfrac{d}{dx}\left(x^4+1\right)\\\\\\\dfrac{du}{dx}~=~(4-1)x^{4-1}\\\\\\\boxed{du~=~\left(4x^3\right)dx}$

$Fazendo~a~substituicao~na~integral~ \displaystyle\int\limits^{1}_{0}~4x^3(x^4+1)^6dx~~,~temos:$  

$\displaystyle\int\limits^{1}_{0}~(u)^6du~=\\\\\\=~\left(~\dfrac{u^7}{7}~\right|_{0}^{1}\\\\\\Voltando~a~substituicao\\\\\\=~\left(\dfrac{(x^4+1)^7}{7}\right|_{0}^{1}\\\\\\=~\dfrac{(1^4+1)^7}{7}~-~\dfrac{(0^4+1)^7}{7}\\\\\\=~\dfrac{2^7}{7}~-~\dfrac{1^7}{7}\\\\\\=~\dfrac{128-1}{7}\\\\\\=~\boxed{\dfrac{127}{7}}$