What is the integral of 4x^2(8x^3+1)^{12} ?

The integral of 4x^2(8x^3+1)^{12} is 1/78 (8x^3+1)^{13}+C

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integral of 4x^2(8x^3+1)^{12}

\int 4 x^{2} (8 x^{3} + 1)^{12} d x

\frac{1}{78} (8 x^{3} + 1)^{13} + C

Solution steps

\int 4 x^{2} (8 x^{3} + 1)^{12} d x

Take the constant out:

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

= 4 \cdot \int x^{2} (8 x^{3} + 1)^{12} d x

Apply u-substitution

: \int \frac{u ( 8 u ^{\frac{3}{2}} + 1 ) ^{12}}{2 x} d u

= 4 \cdot \int \frac{( 8 u + 1 ) ^{12}}{3} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int (8 u + 1)^{12} d u

Apply u-substitution

: \int \frac{v ^{12}}{8} d v

= 4 \cdot \frac{1}{3} \cdot \int \frac{v ^{12}}{8} d v

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \frac{1}{8} \cdot \int v^{12} d v

Apply the Power Rule

: \frac{v ^{13}}{13}

= 4 \cdot \frac{1}{3} \cdot \frac{1}{8} \cdot \frac{v ^{13}}{13}

Substitute back

= 4 \cdot \frac{1}{3} \cdot \frac{1}{8} \cdot \frac{( 8 x ^{3} + 1 ) ^{13}}{13}

Simplify

4 \cdot \frac{1}{3} \cdot \frac{1}{8} \cdot \frac{( 8 x ^{3} + 1 ) ^{13}}{13} : \frac{1}{78} (8 x^{3} + 1)^{13}

= \frac{1}{78} (8 x^{3} + 1)^{13}

Add a constant to the solution

= \frac{1}{78} (8 x^{3} + 1)^{13} + C

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What is the integral of 4x^2(8x^3+1)^{12} ?
The integral of 4x^2(8x^3+1)^{12} is 1/78 (8x^3+1)^{13}+C