Question

Calcule a integral por substituição de variável

Answer 1

$\displaystyle I=\int\! 2x(x^2+1)^{23}\,dx\\\\\\ =\int\! (x^2+1)^{23}\cdot 2x\,dx~~~~~~\mathbf{(i)}$


Faça a seguinte substituição:

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


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

$\displaystyle=\int\! u^{23}\,du\\\\\\ =\frac{u^{23+1}}{23+1}+C\\\\\\ =\frac{u^{24}}{24}+C\\\\\\ =\frac{(x^2+1)^{24}}{24}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\! 2x(x^2+1)^{23}\,dx=\frac{1}{24}\,(x^2+1)^{24}+C \end{array}}$


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \:2x\left(x^2+1\right)^{23}dx\\\\\\{Remova\:a\:constante}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\\\\\=2\cdot \int \:x\left(x^2+1\right)^{23}dx\\\\\\=2\cdot \int \frac{u^{23}}{2}du\\\\\\=2\cdot \int \frac{u^{23}}{2}du\\\\\\{Aplique \:a\:regra\:da\:potência}\to:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\\\=2\cdot \frac{1}{2}\cdot \frac{u^{23+1}}{23+1}\\\\\\=2\cdot \frac{1}{2}\cdot \frac{\left(x^2+1\right)^{23+1}}{23+1}$

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