Me ajudemmmmmmmmmmmmmmm.... Pergunta de ideia deDarlis

October 2020 1 113 Report

Me ajudemmmmmmmmmmmmmmm

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adrielcavalcant Faz o seguinte ...
$\int (8x - 3)^{7}dx = \\\\
u = 8x -3 \\\\
du =8 dx \to dx = \frac{du}{8} \\\\
\int (u)^{7}\frac{du}{8} = \\\\
 \frac{1}{8}\int u^{7}du = \\\\
\frac{1}{8} \frac{u^{7 + 1}}{7+1} + c = \frac{1}{8} \frac{u^{8}}{8}+c = \\\\
 \boxed{\frac{1}{64} u^{8} + c}=>\boxed{\boxed{\frac{1}{64} (8x - 3)^{8} + c}}$
Letra (c)

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