Question

Calcule a integral por substituição de variável

Answer 1

$\displaystyle I=\int\!(3x^2+1)^3\,x\,dx\\\\\\ =\int\!\frac{1}{6}\cdot 6\cdot (3x^2+1)^3\,x\,dx\\\\\\ =\frac{1}{6}\int\!(3x^2+1)^3\cdot 6x\,dx~~~~~~\mathbf{(i)}$


Faça a seguinte substituição:

$3x^2+1=u~~\Rightarrow~~6x\,dx=du$


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

$\displaystyle =\frac{1}{6}\int\!u^3\,du\\\\\\ =\frac{1}{6}\cdot \dfrac{u^{3+1}}{3+1}+C\\\\\\ =\frac{1}{6}\cdot \dfrac{u^4}{4}+C\\\\\\ =\frac{1}{24}\,u^4+C\\\\\\ =\frac{1}{24}\,(3x^2+1)^4+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!(3x^2+1)^3\,x\,dx=\frac{1}{24}\,(3x^2+1)^4+C \end{array}}$


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \left(3x^2+1\right)^3xdx\\\\\\=\int \frac{u^3}{6}du\\\\\\=\frac{1}{6}\cdot \int \:u^3du\\\\\\{Aplique\:a\:regra\:da\:potência}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\\\=\frac{1}{6}\cdot \frac{u^{3+1}}{3+1}\\\\\\=\frac{1}{6}\cdot \frac{\left(3x^2+1\right)^{3+1}}{3+1}\\\\\\=\frac{1}{24}\left(3x^2+1\right)^4$

$\to \boxed{\sf =\frac{1}{24}\left(3x^2+1\right)^4+C}$