What is the integral of 32(8x+7)^3 ?

The integral of 32(8x+7)^3 is (8x+7)^4+C

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integral of 32(8x+7)^3

\int 32 (8 x + 7)^{3} d x

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

Solution steps

\int 32 (8 x + 7)^{3} d x

Take the constant out:

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

= 32 \cdot \int (8 x + 7)^{3} d x

Apply u-substitution

: \int \frac{u ^{3}}{8} d u

= 32 \cdot \int \frac{u ^{3}}{8} d u

Take the constant out:

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

= 32 \cdot \frac{1}{8} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= 32 \cdot \frac{1}{8} \cdot \frac{u ^{4}}{4}

Substitute back

u = 8 x + 7

= 32 \cdot \frac{1}{8} \cdot \frac{( 8 x + 7 ) ^{4}}{4}

Simplify

32 \cdot \frac{1}{8} \cdot \frac{( 8 x + 7 ) ^{4}}{4} : (8 x + 7)^{4}

= (8 x + 7)^{4}

Add a constant to the solution

= (8 x + 7)^{4} + C

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