What is the integral of y^2(4-y^3)^{2/3} ?

The integral of y^2(4-y^3)^{2/3} is -1/5 (4-y^3)^{5/3}+C

integral of y^2(4-y^3)^{2/3}

\int y^{2} (4 - y^{3})^{\frac{2}{3}} d y

- \frac{1}{5} (4 - y^{3})^{\frac{5}{3}} + C

Solution steps

\int y^{2} (4 - y^{3})^{\frac{2}{3}} d y

Apply u-substitution

= \int - \frac{u ^{\frac{2}{3}}}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{3} \cdot \int u^{\frac{2}{3}} d u

Apply the Power Rule

= - \frac{1}{3} \cdot \frac{3}{5} u^{\frac{5}{3}}

Substitute back

u = 4 - y^{3}

= - \frac{1}{3} \cdot \frac{3}{5} (4 - y^{3})^{\frac{5}{3}}

Simplify

- \frac{1}{3} \cdot \frac{3}{5} (4 - y^{3})^{\frac{5}{3}} : - \frac{1}{5} (4 - y^{3})^{\frac{5}{3}}

= - \frac{1}{5} (4 - y^{3})^{\frac{5}{3}}

Add a constant to the solution

= - \frac{1}{5} (4 - y^{3})^{\frac{5}{3}} + C

\frac{\partial}{\partial x} (x^{2} \frac{y}{x ^{4} + y ^{2}})

x^{4} + x^{3}, x^{4} + x^{2} + 6 x

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (\frac{9 - x ^{2}}{8 x})

n = 1 \sum \infty \frac{( x ^{n} )}{7 n}

What is the integral of y^2(4-y^3)^{2/3} ?
The integral of y^2(4-y^3)^{2/3} is -1/5 (4-y^3)^{5/3}+C