What is the integral of (x^2-6x+8)^8(x-3) ?

The integral of (x^2-6x+8)^8(x-3) is 1/18 (x^2-6x+8)^9+C

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integral of (x^2-6x+8)^8(x-3)

\int (x^{2} - 6 x + 8)^{8} (x - 3) d x

\frac{1}{18} (x^{2} - 6 x + 8)^{9} + C

Solution steps

\int (x^{2} - 6 x + 8)^{8} (x - 3) d x

Apply u-substitution

: \int u (u^{2} - 1)^{8} d u

= \int u (u^{2} - 1)^{8} d u

Apply u-substitution

: \int \frac{v ^{8}}{2} d v

= \int \frac{v ^{8}}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int v^{8} d v

Apply the Power Rule

: \frac{v ^{9}}{9}

= \frac{1}{2} \cdot \frac{v ^{9}}{9}

Substitute back

= \frac{1}{2} \cdot \frac{( ( x - 3 ) ^{2} - 1 ) ^{9}}{9}

Simplify

\frac{1}{2} \cdot \frac{( ( x - 3 ) ^{2} - 1 ) ^{9}}{9} : \frac{1}{18} (x^{2} - 6 x + 8)^{9}

= \frac{1}{18} (x^{2} - 6 x + 8)^{9}

Add a constant to the solution

= \frac{1}{18} (x^{2} - 6 x + 8)^{9} + C

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What is the integral of (x^2-6x+8)^8(x-3) ?
The integral of (x^2-6x+8)^8(x-3) is 1/18 (x^2-6x+8)^9+C