What is the integral of (3-8x)^{12} ?

The integral of (3-8x)^{12} is 1/104 (-3+8x)^{13}+C

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integral of (3-8x)^{12}

\int (3 - 8 x)^{12} d x

\frac{1}{104} (- 3 + 8 x)^{13} + C

Solution steps

\int (3 - 8 x)^{12} d x

Factor

3 - 8 x : - (- 3 + 8 x)

= \int (- (- 3 + 8 x))^{12} d x

Simplify

(- (- 3 + 8 x))^{12} : (- 3 + 8 x)^{12}

= \int (- 3 + 8 x)^{12} d x

Apply u-substitution

: \int \frac{u ^{12}}{8} d u

= \int \frac{u ^{12}}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u^{12} d u

Apply the Power Rule

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

= \frac{1}{8} \cdot \frac{u ^{13}}{13}

Substitute back

u = - 3 + 8 x

= \frac{1}{8} \cdot \frac{( - 3 + 8 x ) ^{13}}{13}

Simplify

\frac{1}{8} \cdot \frac{( - 3 + 8 x ) ^{13}}{13} : \frac{1}{104} (- 3 + 8 x)^{13}

= \frac{1}{104} (- 3 + 8 x)^{13}

Add a constant to the solution

= \frac{1}{104} (- 3 + 8 x)^{13} + C

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What is the integral of (3-8x)^{12} ?
The integral of (3-8x)^{12} is 1/104 (-3+8x)^{13}+C