What is the integral of (3x+10)^{12} ?

The integral of (3x+10)^{12} is 1/39 (3x+10)^{13}+C

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integral of (3x+10)^{12}

\int (3 x + 10)^{12} d x

\frac{1}{39} (3 x + 10)^{13} + C

Solution steps

\int (3 x + 10)^{12} d x

Apply u-substitution

: \int \frac{u ^{12}}{3} d u

= \int \frac{u ^{12}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u^{12} d u

Apply the Power Rule

: \frac{u ^{13}}{13}

= \frac{1}{3} \cdot \frac{u ^{13}}{13}

Substitute back

u = 3 x + 10

= \frac{1}{3} \cdot \frac{( 3 x + 10 ) ^{13}}{13}

Simplify

\frac{1}{3} \cdot \frac{( 3 x + 10 ) ^{13}}{13} : \frac{1}{39} (3 x + 10)^{13}

= \frac{1}{39} (3 x + 10)^{13}

Add a constant to the solution

= \frac{1}{39} (3 x + 10)^{13} + C

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What is the integral of (3x+10)^{12} ?
The integral of (3x+10)^{12} is 1/39 (3x+10)^{13}+C