What is the integral of (9-4x)^{12} ?

The integral of (9-4x)^{12} is 1/52 (-9+4x)^{13}+C

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integral of (9-4x)^{12}

\int (9 - 4 x)^{12} d x

\frac{1}{52} (- 9 + 4 x)^{13} + C

Solution steps

\int (9 - 4 x)^{12} d x

Factor

9 - 4 x : - (- 9 + 4 x)

= \int (- (- 9 + 4 x))^{12} d x

Simplify

(- (- 9 + 4 x))^{12} : (- 9 + 4 x)^{12}

= \int (- 9 + 4 x)^{12} d x

Apply u-substitution

= \int \frac{u ^{12}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{12} d u

Apply the Power Rule

= \frac{1}{4} \cdot \frac{u ^{13}}{13}

Substitute back

u = - 9 + 4 x

= \frac{1}{4} \cdot \frac{( - 9 + 4 x ) ^{13}}{13}

Simplify

\frac{1}{4} \cdot \frac{( - 9 + 4 x ) ^{13}}{13} : \frac{1}{52} (- 9 + 4 x)^{13}

= \frac{1}{52} (- 9 + 4 x)^{13}

Add a constant to the solution

= \frac{1}{52} (- 9 + 4 x)^{13} + C

What is the integral of (9-4x)^{12} ?
The integral of (9-4x)^{12} is 1/52 (-9+4x)^{13}+C