Question

Calcule a integral por substituição de variável

Answer 1

$\displaystyle I=\int\!(x^3+4)^3\,x^2\,dx\\\\\\ =\int\!\frac{1}{3}\cdot 3(x^3+4)^3\,x^2\,dx\\\\\\ =\frac{1}{3}\int\!(x^3+4)^3\cdot 3x^2\,dx~~~~~~\mathbf{(i)}$


Faça a seguinte substituição:

$x^3+4=u~~\Rightarrow~~3x^2\,dx=du$


Substituindo em $\mathbf{(i)},$ a integral fica

$\displaystyle=\frac{1}{3}\int\! u^3\,du\\\\\\ =\frac{1}{3}\cdot \dfrac{u^{3+1}}{3+1}+C\\\\\\ =\frac{1}{3}\cdot \dfrac{u^4}{4}+C\\\\\\ =\frac{1}{12}\,u^4+C\\\\\\ =\frac{1}{12}\,(x^3+4)^4+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c}\displaystyle\int\!(x^3+4)^3\,x^2\,dx=\frac{1}{12}\,(x^3+4)^4+C \end{array}}$


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \left(x^3+4\right)^3\cdot x^2dx\\\\\\=\int \:x^{11}+12x^8+48x^5+64x^2dx\\\\\\=\int \:x^{11}dx+\int \:12x^8dx+\int \:48x^5dx+\int \:64x^2dx\\\\\\=\frac{x^{12}}{12}+\frac{4x^9}{3}+8x^6+\frac{64x^3}{3}\\\\\\\to \boxed{\sf =\frac{x^{12}}{12}+\frac{4x^9}{3}+8x^6+\frac{64x^3}{3}+C}$