What is the integral of 5(9x-4)^4 ?

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

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integral of 5(9x-4)^4

\int 5 (9 x - 4)^{4} d x

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

Solution steps

\int 5 (9 x - 4)^{4} d x

Take the constant out:

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

= 5 \cdot \int (9 x - 4)^{4} d x

Apply u-substitution

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

= 5 \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

= 5 \cdot \frac{1}{9} \cdot \int u^{4} d u

Apply the Power Rule

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

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

Substitute back

u = 9 x - 4

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

Simplify

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

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

Add a constant to the solution

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

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