What is the integral of y(1-y)^{1/3} ?

The integral of y(1-y)^{1/3} is (1-y)^{1/3}yx+C

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integral of y(1-y)^{1/3}

\int y (1 - y)^{\frac{1}{3}} d x

(1 - y)^{\frac{1}{3}} y x + C

Solution steps

\int y (1 - y)^{\frac{1}{3}} d x

Integral of a constant:

\int a d x = a x

= (1 - y)^{\frac{1}{3}} y x

Add a constant to the solution

= (1 - y)^{\frac{1}{3}} y x + C

\int sec^{2} (x) \cdot (1 + tan (x)) d x

\int 0.5 e^{0.1 x} d x

\int tan (1 + π t - t) d t

\int \frac{10 x ^{3}}{x ^{2} + 9} d x

\int - 11

What is the integral of y(1-y)^{1/3} ?
The integral of y(1-y)^{1/3} is (1-y)^{1/3}yx+C