What is the integral of x^2(3x^3+14)^3 ?

The integral of x^2(3x^3+14)^3 is 1/36 (3x^3+14)^4+C

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integral of x^2(3x^3+14)^3

\int x^{2} (3 x^{3} + 14)^{3} d x

\frac{1}{36} (3 x^{3} + 14)^{4} + C

Solution steps

\int x^{2} (3 x^{3} + 14)^{3} d x

Apply u-substitution

: \int \frac{u ( 3 u ^{\frac{3}{2}} + 14 ) ^{3}}{2 x} d u

= \int \frac{( 3 u + 14 ) ^{3}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int (3 u + 14)^{3} d u

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

= \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{( 3 x ^{3} + 14 ) ^{4}}{4}

Simplify

\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{( 3 x ^{3} + 14 ) ^{4}}{4} : \frac{1}{36} (3 x^{3} + 14)^{4}

= \frac{1}{36} (3 x^{3} + 14)^{4}

Add a constant to the solution

= \frac{1}{36} (3 x^{3} + 14)^{4} + C

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What is the integral of x^2(3x^3+14)^3 ?
The integral of x^2(3x^3+14)^3 is 1/36 (3x^3+14)^4+C