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

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

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integral of 21(x+4)^6

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

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

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = x + 4

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

Simplify

21 \cdot \frac{( 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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