What is the integral of 15(2x+4)^4 ?

The integral of 15(2x+4)^4 is 3/2 (2x+4)^5+C

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integral of 15(2x+4)^4

\int 15 (2 x + 4)^{4} d x

\frac{3}{2} (2 x + 4)^{5} + C

Solution steps

\int 15 (2 x + 4)^{4} d x

Take the constant out:

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

= 15 \cdot \int (2 x + 4)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{2} d u

= 15 \cdot \int \frac{u ^{4}}{2} d u

Take the constant out:

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

= 15 \cdot \frac{1}{2} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= 15 \cdot \frac{1}{2} \cdot \frac{u ^{5}}{5}

Substitute back

u = 2 x + 4

= 15 \cdot \frac{1}{2} \cdot \frac{( 2 x + 4 ) ^{5}}{5}

Simplify

15 \cdot \frac{1}{2} \cdot \frac{( 2 x + 4 ) ^{5}}{5} : \frac{3}{2} (2 x + 4)^{5}

= \frac{3}{2} (2 x + 4)^{5}

Add a constant to the solution

= \frac{3}{2} (2 x + 4)^{5} + C

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What is the integral of 15(2x+4)^4 ?
The integral of 15(2x+4)^4 is 3/2 (2x+4)^5+C