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

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

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integral of (3-2x)^2

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

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

Solution steps

\int (3 - 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 = 3 - 2 x

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

Simplify

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

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

Add a constant to the solution

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

f (x) = \frac{1}{4} x^{8}

\frac{d}{d x} (\frac{1 - sin ( x )}{1 - cos ( x )})

t^{3} cos (t)

\int \frac{1}{x ^{4} + x} d x

x \to 0 lim (\frac{( 2 - x ) ^{2} - 4}{x})