What is the integral of 9z^2(27z^3+1)^{12} ?

The integral of 9z^2(27z^3+1)^{12} is 1/117 (27z^3+1)^{13}+C

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integral of 9z^2(27z^3+1)^{12}

\int 9 z^{2} (27 z^{3} + 1)^{12} d z

\frac{1}{117} (27 z^{3} + 1)^{13} + C

Solution steps

\int 9 z^{2} (27 z^{3} + 1)^{12} d z

Take the constant out:

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

= 9 \cdot \int z^{2} (27 z^{3} + 1)^{12} d z

Apply u-substitution

: \int \frac{u ( 27 u ^{\frac{3}{2}} + 1 ) ^{12}}{2 z} d u

= 9 \cdot \int \frac{( 27 u + 1 ) ^{12}}{3} d u

Take the constant out:

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

= 9 \cdot \frac{1}{3} \cdot \int (27 u + 1)^{12} d u

Apply u-substitution

: \int \frac{v ^{12}}{27} d v

= 9 \cdot \frac{1}{3} \cdot \int \frac{v ^{12}}{27} d v

Take the constant out:

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

= 9 \cdot \frac{1}{3} \cdot \frac{1}{27} \cdot \int v^{12} d v

Apply the Power Rule

: \frac{v ^{13}}{13}

= 9 \cdot \frac{1}{3} \cdot \frac{1}{27} \cdot \frac{v ^{13}}{13}

Substitute back

= 9 \cdot \frac{1}{3} \cdot \frac{1}{27} \cdot \frac{( 27 z ^{3} + 1 ) ^{13}}{13}

Simplify

9 \cdot \frac{1}{3} \cdot \frac{1}{27} \cdot \frac{( 27 z ^{3} + 1 ) ^{13}}{13} : \frac{1}{117} (27 z^{3} + 1)^{13}

= \frac{1}{117} (27 z^{3} + 1)^{13}

Add a constant to the solution

= \frac{1}{117} (27 z^{3} + 1)^{13} + C

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What is the integral of 9z^2(27z^3+1)^{12} ?
The integral of 9z^2(27z^3+1)^{12} is 1/117 (27z^3+1)^{13}+C