What is the integral of 3(3x+9)^6 ?

The integral of 3(3x+9)^6 is 1/7 (3x+9)^7+C

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integral of 3(3x+9)^6

\int 3 (3 x + 9)^{6} d x

\frac{1}{7} (3 x + 9)^{7} + C

Solution steps

\int 3 (3 x + 9)^{6} 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)^{6} d x

Apply u-substitution

: \int \frac{u ^{6}}{3} d u

= 3 \cdot \int \frac{u ^{6}}{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^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

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

Substitute back

u = 3 x + 9

= 3 \cdot \frac{1}{3} \cdot \frac{( 3 x + 9 ) ^{7}}{7}

Simplify

3 \cdot \frac{1}{3} \cdot \frac{( 3 x + 9 ) ^{7}}{7} : \frac{1}{7} (3 x + 9)^{7}

= \frac{1}{7} (3 x + 9)^{7}

Add a constant to the solution

= \frac{1}{7} (3 x + 9)^{7} + C

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