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

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

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integral of 3(3x+9)^3

\int 3 (3 x + 9)^{3} d x

\frac{1}{4} (3 x + 9)^{4} + C

Solution steps

\int 3 (3 x + 9)^{3} d x

Take the constant out:

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

= 3 \cdot \int (3 x + 9)^{3} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

= 3 \cdot \frac{1}{3} \cdot \frac{u ^{4}}{4}

Substitute back

u = 3 x + 9

= 3 \cdot \frac{1}{3} \cdot \frac{( 3 x + 9 ) ^{4}}{4}

Simplify

3 \cdot \frac{1}{3} \cdot \frac{( 3 x + 9 ) ^{4}}{4} : \frac{1}{4} (3 x + 9)^{4}

= \frac{1}{4} (3 x + 9)^{4}

Add a constant to the solution

= \frac{1}{4} (3 x + 9)^{4} + C

x \to 0 lim (\frac{sinh ( x ) - x}{x ^{3}})

e^{n \cdot x}

\frac{x ^{2}}{4}, \frac{- x + 6}{4}, - 3, 2

\frac{10 x + 2 x ^{2}}{( 5 + 2 x ) ^{2}}

\frac{d y}{d t} + a y + b = 0