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

The integral of (9x+4)^4 is 1/45 (9x+4)^5+C

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integral of (9x+4)^4

integral

(9 x + 4)^{4}

\frac{1}{45} (9 x + 4)^{5} + C

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

= \frac{1}{9} \cdot \frac{u ^{5}}{5}

Substitute back

u = 9 x + 4

= \frac{1}{9} \cdot \frac{( 9 x + 4 ) ^{5}}{5}

Simplify

\frac{1}{9} \cdot \frac{( 9 x + 4 ) ^{5}}{5} : \frac{1}{45} (9 x + 4)^{5}

= \frac{1}{45} (9 x + 4)^{5}

Add a constant to the solution

= \frac{1}{45} (9 x + 4)^{5} + C