Question

integral((raiz cubica de x² )+1)dx

Answer 1

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Calcular a integral indefinida:

$\large\begin{array}{l} \mathsf{\displaystyle\int\!\big(\,^3\!\!\!\sqrt{x^2}+1\big)\,dx=\int\!\big(x^{\frac{\,2\,}{3}}+1\big)\,dx}\\\\ \mathsf{\displaystyle\int\!\big(\,^3\!\!\!\sqrt{x^2}+1\big)\,dx=\int\! x^{\frac{\,2\,}{3}}\,dx+\int\! 1\,dx}\\\\ \mathsf{\displaystyle\int\!\big(\,^3\!\!\!\sqrt{x^2}+1\big)\,dx=\frac{x^{\frac{\,2\,}{3}+1}}{\frac{\,2\,}{3}+1}+x+C} \end{array}$

$\large\begin{array}{l} \mathsf{\displaystyle\int\!\big(\,^3\!\!\!\sqrt{x^2}+1\big)\,dx=\frac{x^{\frac{\,5\,}{3}}}{\frac{\,5\,}{3}}+x+C}\\\\ \mathsf{\displaystyle\int\!\big(\,^3\!\!\!\sqrt{x^2}+1\big)\,dx=\frac{\,3\,}{5}\,x^{\frac{\,5\,}{3}}+x+C}\\\\\\ \therefore~~\mathsf{\displaystyle\int\!\big(\,^3\!\!\!\sqrt{x^2}+1\big)\,dx=\frac{\,3\,}{5}\,^3\!\!\!\sqrt{x^5}+x+C}\quad\longleftarrow\quad\textsf{esta \'e a resposta.} \end{array}$


Bons estudos! :-)


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