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

The integral of (6x^2-4x+2)(x^3-x^2+x+1)^7 is 1/4 (x^3-x^2+x+1)^8+C

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integral of (6x^2-4x+2)(x^3-x^2+x+1)^7

\int (6 x^{2} - 4 x + 2) (x^{3} - x^{2} + x + 1)^{7} d x

\frac{1}{4} (x^{3} - x^{2} + x + 1)^{8} + C

Solution steps

\int (6 x^{2} - 4 x + 2) (x^{3} - x^{2} + x + 1)^{7} d x

Apply u-substitution

: \int 2 u^{7} d u

= \int 2 u^{7} d u

Take the constant out:

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

= 2 \cdot \int u^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= 2 \cdot \frac{u ^{8}}{8}

Substitute back

u = x^{3} - x^{2} + x + 1

= 2 \cdot \frac{( x ^{3} - x ^{2} + x + 1 ) ^{8}}{8}

Simplify

2 \cdot \frac{( x ^{3} - x ^{2} + x + 1 ) ^{8}}{8} : \frac{1}{4} (x^{3} - x^{2} + x + 1)^{8}

= \frac{1}{4} (x^{3} - x^{2} + x + 1)^{8}

Add a constant to the solution

= \frac{1}{4} (x^{3} - x^{2} + x + 1)^{8} + C

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What is the integral of (6x^2-4x+2)(x^3-x^2+x+1)^7 ?
The integral of (6x^2-4x+2)(x^3-x^2+x+1)^7 is 1/4 (x^3-x^2+x+1)^8+C