What is the integral of 6(x-1)^2 ?

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

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integral of 6(x-1)^2

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

6 (\frac{x ^{3}}{3} - x^{2} + x) + C

Solution steps

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

Take the constant out:

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

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

Expand

(x - 1)^{2} : x^{2} - 2 x + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int x^{2} d x = \frac{x ^{3}}{3}

\int 2 x d x = x^{2}

\int 1 d x = x

= 6 (\frac{x ^{3}}{3} - x^{2} + x)

Add a constant to the solution

= 6 (\frac{x ^{3}}{3} - x^{2} + x) + C

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