What is the integral of 3(3x+9)^7 ?

The integral of 3(3x+9)^7 is 1/8 (3x+9)^8+C

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integral of 3(3x+9)^7

\int 3 (3 x + 9)^{7} d x

\frac{1}{8} (3 x + 9)^{8} + C

Solution steps

\int 3 (3 x + 9)^{7} d x

Take the constant out:

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

= 3 \cdot \int (3 x + 9)^{7} d x

Apply u-substitution

= 3 \cdot \int \frac{u ^{7}}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int u^{7} d u

Apply the Power Rule

= 3 \cdot \frac{1}{3} \cdot \frac{u ^{8}}{8}

Substitute back

u = 3 x + 9

= 3 \cdot \frac{1}{3} \cdot \frac{( 3 x + 9 ) ^{8}}{8}

Simplify

3 \cdot \frac{1}{3} \cdot \frac{( 3 x + 9 ) ^{8}}{8} : \frac{1}{8} (3 x + 9)^{8}

= \frac{1}{8} (3 x + 9)^{8}

Add a constant to the solution

= \frac{1}{8} (3 x + 9)^{8} + C