What is the integral of (4-2x)^3 ?

The integral of (4-2x)^3 is -1/8 (4-2x)^4+C

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integral of (4-2x)^3

integral

(4 - 2 x)^{3}

- \frac{1}{8} (4 - 2 x)^{4} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: \frac{u ^{4}}{4}

= - \frac{1}{2} \cdot \frac{u ^{4}}{4}

Substitute back

u = 4 - 2 x

= - \frac{1}{2} \cdot \frac{( 4 - 2 x ) ^{4}}{4}

Simplify

- \frac{1}{2} \cdot \frac{( 4 - 2 x ) ^{4}}{4} : - \frac{1}{8} (4 - 2 x)^{4}

= - \frac{1}{8} (4 - 2 x)^{4}

Add a constant to the solution

= - \frac{1}{8} (4 - 2 x)^{4} + C

\frac{3}{x ^{3}}

\frac{1}{a ^{2} e ^{x} + e ^{- x}}

y + 1

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

\frac{x}{x ^{2} + y ^{2}}