Matemática, perguntado por DSM30, 1 ano atrás

∫ 16x(4x²+2)^7 dx é igual

Soluções para a tarefa

Respondido por Usuário anônimo
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Boa tarde !

Solução!

Integral por substituição!

\displaystyle \int 16x(4 x^{2} +2)^{7}dx\\\\\\\\ 16 \displaystyle \int x(4 x^{2} +2)^{7}dx\\\\\\\\

u=(4 x^{2} +2)~~~~du=8xdx= \dfrac{du}{8x}=dx\\\\\\\\\\ 16\displaystyle \int \dfrac{x(u)^{7} }{8x}du\\\\\\\\\\\ 16\displaystyle \int \dfrac{(u)^{7} }{8}du\\\\\\\\\\ 16. \frac{1}{8} \displaystyle \int {(u)^{7} du


\dfrac{16}{8} \left (\dfrac{(u)^{(7+1)} }{(7+1)} \right )\\\\\\\\\ \dfrac{16}{8} \left (\dfrac{(u)^{8} }{8} \right )\\\\\\\\\\ 2. \left (\dfrac{(u)^{8} }{8} \right )\\\\\\\\\\  \left (\dfrac{2.(u)^{8} }{8} \right )\\\\\\\\\ \left (\dfrac{(u)^{8} }{4} \right )\\\\\\\\\  \left (\dfrac{(4 x^{2} +2)^{8} }{4}+c \right )


 \boxed{Resposta:\displaystyle \int 16x(4 x^{2} +2)^{7}dx=\left (\dfrac{(4 x^{2} +2)^{8} }{4}+c \right )}

Boa tarde!
Bons estudos!


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