What is the integral of x^2(4x^3-3)^4 ?

The integral of x^2(4x^3-3)^4 is 1/60 (4x^3-3)^5+C

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integral of x^2(4x^3-3)^4

\int x^{2} (4 x^{3} - 3)^{4} d x

\frac{1}{60} (4 x^{3} - 3)^{5} + C

Solution steps

\int x^{2} (4 x^{3} - 3)^{4} d x

Apply u-substitution

: \int \frac{u ( 4 u ^{\frac{3}{2}} - 3 ) ^{4}}{2 x} d u

= \int \frac{( 4 u - 3 ) ^{4}}{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 (4 u - 3)^{4} d u

Apply u-substitution

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

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

Apply the Power Rule

: \frac{v ^{5}}{5}

= \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{v ^{5}}{5}

Substitute back

= \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{( 4 x ^{3} - 3 ) ^{5}}{5}

Simplify

\frac{1}{3} \cdot \frac{1}{4} \cdot \frac{( 4 x ^{3} - 3 ) ^{5}}{5} : \frac{1}{60} (4 x^{3} - 3)^{5}

= \frac{1}{60} (4 x^{3} - 3)^{5}

Add a constant to the solution

= \frac{1}{60} (4 x^{3} - 3)^{5} + C

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What is the integral of x^2(4x^3-3)^4 ?
The integral of x^2(4x^3-3)^4 is 1/60 (4x^3-3)^5+C