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

The integral of (x^2-2x+6)(2x-2) is 1/2 (x^2-2x+6)^2+C

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

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

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

Solution steps

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

(x^{2} - 2 x + 6) (2 x - 2) = 2 (x^{2} - 2 x + 6) (x - 1)

= \int 2 (x^{2} - 2 x + 6) (x - 1) d x

Take the constant out:

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

= 2 \cdot \int (x^{2} - 2 x + 6) (x - 1) d x

Apply u-substitution

: \int \frac{u}{2} d u

= 2 \cdot \int \frac{u}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{2}}{2}

Substitute back

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

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 2 x + 6 ) ^{2}}{2}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 2 x + 6 ) ^{2}}{2} : \frac{1}{2} (x^{2} - 2 x + 6)^{2}

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

Add a constant to the solution

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

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