What is the integral of 18(3x+4)^5 ?

The integral of 18(3x+4)^5 is (3x+4)^6+C

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integral of 18(3x+4)^5

\int 18 (3 x + 4)^{5} d x

(3 x + 4)^{6} + C

Solution steps

\int 18 (3 x + 4)^{5} d x

Take the constant out:

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

= 18 \cdot \int (3 x + 4)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{3} d u

= 18 \cdot \int \frac{u ^{5}}{3} d u

Take the constant out:

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

= 18 \cdot \frac{1}{3} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= 18 \cdot \frac{1}{3} \cdot \frac{u ^{6}}{6}

Substitute back

u = 3 x + 4

= 18 \cdot \frac{1}{3} \cdot \frac{( 3 x + 4 ) ^{6}}{6}

Simplify

18 \cdot \frac{1}{3} \cdot \frac{( 3 x + 4 ) ^{6}}{6} : (3 x + 4)^{6}

= (3 x + 4)^{6}

Add a constant to the solution

= (3 x + 4)^{6} + C

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