What is the integral of (3x+9)^4 ?

The integral of (3x+9)^4 is 1/15 (3x+9)^5+C

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integral of (3x+9)^4

integral

(3 x + 9)^{4}

\frac{1}{15} (3 x + 9)^{5} + C

Solution steps

\int (3 x + 9)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{3} d u

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

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{1}{3} \cdot \frac{u ^{5}}{5}

Substitute back

u = 3 x + 9

= \frac{1}{3} \cdot \frac{( 3 x + 9 ) ^{5}}{5}

Simplify

\frac{1}{3} \cdot \frac{( 3 x + 9 ) ^{5}}{5} : \frac{1}{15} (3 x + 9)^{5}

= \frac{1}{15} (3 x + 9)^{5}

Add a constant to the solution

= \frac{1}{15} (3 x + 9)^{5} + C

\frac{( x + 1 )}{( x ^{2} + x + 1 )}

ln (\frac{e ^{x}}{x ^{2}})

(x^{2} - a^{2})^{\frac{1}{2}}

6 e^{- 2} x

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