Answer to Question #348854 in Calculus for Mary Claire Daza

Question #348854

"Solve \t\u222b\u2592\u221bx dx"


1
Expert's answer
2022-06-08T04:57:56-0400

"\\int {\\sqrt[3]{x}} dx = \\int {{x^{\\frac{1}{3}}}} dx = \\frac{{{x^{\\frac{1}{3} + 1}}}}{{\\frac{1}{3} + 1}} + C = \\frac{{{x^{\\frac{4}{3}}}}}{{\\frac{4}{3}}} + C = \\frac{3}{4}x\\sqrt[3]{x} + C"

Answer: "\\frac{3}{4}x\\sqrt[3]{x} + C"


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