What is the integral of (x^2-x)(4x^3-6x^2+3)^6 ?

The integral of (x^2-x)(4x^3-6x^2+3)^6 is 1/84 (4x^3-6x^2+3)^7+C

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integral of (x^2-x)(4x^3-6x^2+3)^6

\int (x^{2} - x) (4 x^{3} - 6 x^{2} + 3)^{6} d x

\frac{1}{84} (4 x^{3} - 6 x^{2} + 3)^{7} + C

Solution steps

\int (x^{2} - x) (4 x^{3} - 6 x^{2} + 3)^{6} d x

Apply u-substitution

: \int \frac{u ^{6}}{12} d u

= \int \frac{u ^{6}}{12} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= \frac{1}{12} \cdot \frac{u ^{7}}{7}

Substitute back

u = 4 x^{3} - 6 x^{2} + 3

= \frac{1}{12} \cdot \frac{( 4 x ^{3} - 6 x ^{2} + 3 ) ^{7}}{7}

Simplify

\frac{1}{12} \cdot \frac{( 4 x ^{3} - 6 x ^{2} + 3 ) ^{7}}{7} : \frac{1}{84} (4 x^{3} - 6 x^{2} + 3)^{7}

= \frac{1}{84} (4 x^{3} - 6 x^{2} + 3)^{7}

Add a constant to the solution

= \frac{1}{84} (4 x^{3} - 6 x^{2} + 3)^{7} + C

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What is the integral of (x^2-x)(4x^3-6x^2+3)^6 ?
The integral of (x^2-x)(4x^3-6x^2+3)^6 is 1/84 (4x^3-6x^2+3)^7+C