What is the integral of 21(3x+4)^6 ?

The integral of 21(3x+4)^6 is (3x+4)^7+C

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integral of 21(3x+4)^6

\int 21 (3 x + 4)^{6} d x

(3 x + 4)^{7} + C

Solution steps

\int 21 (3 x + 4)^{6} d x

Take the constant out:

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

= 21 \cdot \int (3 x + 4)^{6} d x

Apply u-substitution

: \int \frac{u ^{6}}{3} d u

= 21 \cdot \int \frac{u ^{6}}{3} d u

Take the constant out:

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

= 21 \cdot \frac{1}{3} \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= 21 \cdot \frac{1}{3} \cdot \frac{u ^{7}}{7}

Substitute back

u = 3 x + 4

= 21 \cdot \frac{1}{3} \cdot \frac{( 3 x + 4 ) ^{7}}{7}

Simplify

21 \cdot \frac{1}{3} \cdot \frac{( 3 x + 4 ) ^{7}}{7} : (3 x + 4)^{7}

= (3 x + 4)^{7}

Add a constant to the solution

= (3 x + 4)^{7} + C

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