What is the integral of (9x-16)^3 ?

The integral of (9x-16)^3 is 1/36 (9x-16)^4+C

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integral of (9x-16)^3

\int (9 x - 16)^{3} d x

\frac{1}{36} (9 x - 16)^{4} + C

Solution steps

\int (9 x - 16)^{3} d x

Apply u-substitution

: \int \frac{u ^{3}}{9} d u

= \int \frac{u ^{3}}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{1}{9} \cdot \frac{u ^{4}}{4}

Substitute back

u = 9 x - 16

= \frac{1}{9} \cdot \frac{( 9 x - 16 ) ^{4}}{4}

Simplify

\frac{1}{9} \cdot \frac{( 9 x - 16 ) ^{4}}{4} : \frac{1}{36} (9 x - 16)^{4}

= \frac{1}{36} (9 x - 16)^{4}

Add a constant to the solution

= \frac{1}{36} (9 x - 16)^{4} + C

\frac{d}{d x} (ln (\frac{3 x + 4}{3 x - 4}))

\int sin (t) 2 t d t

\int \frac{sin ( 2 x )}{e ^{x}} d x

\frac{d}{d x} (\frac{4 e ^{x}}{2 e ^{x} + 1})

2 sin (2 t)

What is the integral of (9x-16)^3 ?
The integral of (9x-16)^3 is 1/36 (9x-16)^4+C