What is the integral of 3(x-1)^2 ?

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

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integral of 3(x-1)^2

\int 3 (x - 1)^{2} d x

x^{3} - 3 x^{2} + 3 x + C

Solution steps

\int 3 (x - 1)^{2} d x

Take the constant out:

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

= 3 \cdot \int (x - 1)^{2} d x

Expand

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

= 3 \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

= 3 (\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

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

Simplify

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

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

Add a constant to the solution

= x^{3} - 3 x^{2} + 3 x + C

\int 3 x^{2} y + 4 x y^{2} d x

t \to \infty lim (\frac{1}{5 t})

\int_{0}^{1} \frac{1}{x ^{2} - 4} d x

\frac{d}{d y} ((y - 4)^{3})