What is the integral of 8(9+x)^5 ?

The integral of 8(9+x)^5 is 4/3 (x+9)^6+C

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integral of 8(9+x)^5

\int 8 (9 + x)^{5} d x

\frac{4}{3} (x + 9)^{6} + C

Solution steps

\int 8 (9 + x)^{5} d x

Take the constant out:

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

= 8 \cdot \int (9 + x)^{5} d x

Apply u-substitution

: \int u^{5} d u

= 8 \cdot \int u^{5} d u

Apply the Power Rule

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

= 8 \cdot \frac{u ^{6}}{6}

Substitute back

u = 9 + x

= 8 \cdot \frac{( 9 + x ) ^{6}}{6}

Simplify

8 \cdot \frac{( 9 + x ) ^{6}}{6} : \frac{4}{3} (x + 9)^{6}

= \frac{4}{3} (x + 9)^{6}

Add a constant to the solution

= \frac{4}{3} (x + 9)^{6} + C

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