What is the integral of (2x-4)^5 ?

The integral of (2x-4)^5 is 1/12 (2x-4)^6+C

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integral of (2x-4)^5

\int (2 x - 4)^{5} d x

\frac{1}{12} (2 x - 4)^{6} + C

Solution steps

\int (2 x - 4)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{2} d u

= \int \frac{u ^{5}}{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^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = 2 x - 4

= \frac{1}{2} \cdot \frac{( 2 x - 4 ) ^{6}}{6}

Simplify

\frac{1}{2} \cdot \frac{( 2 x - 4 ) ^{6}}{6} : \frac{1}{12} (2 x - 4)^{6}

= \frac{1}{12} (2 x - 4)^{6}

Add a constant to the solution

= \frac{1}{12} (2 x - 4)^{6} + C

\frac{\partial}{\partial x} (4 x^{2} + 3 y^{2} - 4 x y + x)

\frac{d}{d x} (e^{t a n (3 x)})

\frac{d}{d t} (\frac{t}{2})

\int \frac{12}{x ^{3} + 8} d x

t^{3} + 4 t^{2} + 5