What is the integral of 9(9x+2)^4 ?

The integral of 9(9x+2)^4 is 1/5 (9x+2)^5+C

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integral of 9(9x+2)^4

\int 9 (9 x + 2)^{4} d x

\frac{1}{5} (9 x + 2)^{5} + C

Solution steps

\int 9 (9 x + 2)^{4} d x

Take the constant out:

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

= 9 \cdot \int (9 x + 2)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{9} d u

= 9 \cdot \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

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

Apply the Power Rule

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

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

Substitute back

u = 9 x + 2

= 9 \cdot \frac{1}{9} \cdot \frac{( 9 x + 2 ) ^{5}}{5}

Simplify

9 \cdot \frac{1}{9} \cdot \frac{( 9 x + 2 ) ^{5}}{5} : \frac{1}{5} (9 x + 2)^{5}

= \frac{1}{5} (9 x + 2)^{5}

Add a constant to the solution

= \frac{1}{5} (9 x + 2)^{5} + C

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