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

The integral of (3x+4)^6 is 1/21 (3x+4)^7+C

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integral of (3x+4)^6

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Substitute back

u = 3 x + 4

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

Simplify

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

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

Add a constant to the solution

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

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