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

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

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integral of 2(2x+4)^5

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int (2 x + 4)^{5} d x

Apply u-substitution

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

= 2 \cdot \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

= 2 \cdot \frac{1}{2} \cdot \int u^{5} d u

Apply the Power Rule

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

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

Substitute back

u = 2 x + 4

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (4^{4})

\int x^{2} cos (\frac{1}{5} x) d x

f (x) = (4 - x) (\frac{2}{x ^{2}} - 3)

y = \frac{28}{x}

y = - e^{- x - 2}, (- 3, - e)