What is the integral of (2-2x)^2 ?

The integral of (2-2x)^2 is -1/6 (2-2x)^3+C

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integral of (2-2x)^2

\int (2 - 2 x)^{2} d x

- \frac{1}{6} (2 - 2 x)^{3} + C

Solution steps

\int (2 - 2 x)^{2} d x

Apply u-substitution

: \int - \frac{u ^{2}}{2} d u

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{3}}{3}

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

Substitute back

u = 2 - 2 x

= - \frac{1}{2} \cdot \frac{( 2 - 2 x ) ^{3}}{3}

Simplify

- \frac{1}{2} \cdot \frac{( 2 - 2 x ) ^{3}}{3} : - \frac{1}{6} (2 - 2 x)^{3}

= - \frac{1}{6} (2 - 2 x)^{3}

Add a constant to the solution

= - \frac{1}{6} (2 - 2 x)^{3} + C

\frac{( x ^{2} + 32 x )}{8}

y = 1 - \frac{x ^{2}}{36}

\frac{\partial}{\partial z} (2 y + z)

\int \frac{arcsin ( 5 x )}{1 - 25 x ^{2}} d x

t^{2} y^{^{''}} + 11 t y^{^{'}} + 25 y = 0