What is the integral of (2-y^3)^2 ?

The integral of (2-y^3)^2 is 4y-y^4+(y^7)/7+C

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integral of (2-y^3)^2

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

4 y - y^{4} + \frac{y ^{7}}{7} + C

Solution steps

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

Expand

(2 - y^{3})^{2} : 4 - 4 y^{3} + y^{6}

= \int 4 - 4 y^{3} + y^{6} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 4 d y - \int 4 y^{3} d y + \int y^{6} d y

\int 4 d y = 4 y

\int 4 y^{3} d y = y^{4}

\int y^{6} d y = \frac{y ^{7}}{7}

= 4 y - y^{4} + \frac{y ^{7}}{7}

Add a constant to the solution

= 4 y - y^{4} + \frac{y ^{7}}{7} + C

\int \frac{2 x - 1}{x ^{2} - x + 3} d x

\int (\frac{7}{x}) d x

\int (x + 5) (x^{2} + 5 x - 10) d x

\frac{\partial}{\partial x} (e^{y} + x + x^{3} + 6)

\frac{d}{d x} (\frac{tan ( x )}{sin ( x )})

What is the integral of (2-y^3)^2 ?
The integral of (2-y^3)^2 is 4y-y^4+(y^7)/7+C