What is the integral of (7x+6)^4 ?

The integral of (7x+6)^4 is 1/35 (7x+6)^5+C

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integral of (7x+6)^4

\int (7 x + 6)^{4} d x

\frac{1}{35} (7 x + 6)^{5} + C

Solution steps

\int (7 x + 6)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{7} d u

= \int \frac{u ^{4}}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int u^{4} d u

Apply the Power Rule

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

= \frac{1}{7} \cdot \frac{u ^{5}}{5}

Substitute back

u = 7 x + 6

= \frac{1}{7} \cdot \frac{( 7 x + 6 ) ^{5}}{5}

Simplify

\frac{1}{7} \cdot \frac{( 7 x + 6 ) ^{5}}{5} : \frac{1}{35} (7 x + 6)^{5}

= \frac{1}{35} (7 x + 6)^{5}

Add a constant to the solution

= \frac{1}{35} (7 x + 6)^{5} + C

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