What is the integral of (8x^4+5)^332x^3 ?

The integral of (8x^4+5)^332x^3 is 1/4 (8x^4+5)^4+C

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integral of (8x^4+5)^332x^3

\int (8 x^{4} + 5)^{3} 32 x^{3} d x

\frac{1}{4} (8 x^{4} + 5)^{4} + C

Solution steps

\int (8 x^{4} + 5)^{3} \cdot 32 x^{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^{4} + 5)^{3} x^{3} d x

Apply u-substitution

: \int \frac{( 8 u + 5 ) ^{3}}{4} d u

= 32 \cdot \int \frac{( 8 u + 5 ) ^{3}}{4} d u

Take the constant out:

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

= 32 \cdot \frac{1}{4} \cdot \int (8 u + 5)^{3} d u

Apply u-substitution

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

= 32 \cdot \frac{1}{4} \cdot \int \frac{v ^{3}}{8} d v

Take the constant out:

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

= 32 \cdot \frac{1}{4} \cdot \frac{1}{8} \cdot \int v^{3} d v

Apply the Power Rule

: \frac{v ^{4}}{4}

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

Substitute back

= 32 \cdot \frac{1}{4} \cdot \frac{1}{8} \cdot \frac{( 8 x ^{4} + 5 ) ^{4}}{4}

Simplify

32 \cdot \frac{1}{4} \cdot \frac{1}{8} \cdot \frac{( 8 x ^{4} + 5 ) ^{4}}{4} : \frac{1}{4} (8 x^{4} + 5)^{4}

= \frac{1}{4} (8 x^{4} + 5)^{4}

Add a constant to the solution

= \frac{1}{4} (8 x^{4} + 5)^{4} + C

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What is the integral of (8x^4+5)^332x^3 ?
The integral of (8x^4+5)^332x^3 is 1/4 (8x^4+5)^4+C