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

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

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integral of-(x-1)^2

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

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

Solution steps

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

Take the constant out:

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

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

Expand

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

= - \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

= - (\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

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

Simplify

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

Add a constant to the solution

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

f (x) = x^{2} e^{6 x}

\frac{d}{d x} (cos (x) - 4 tan (x))

\frac{d}{d x} (x arctan (x))

2 y^{^{''}} - y^{^{'}} + 3 y = 0

x^{2} y^{^{''}} - 14 x y^{^{'}} + 56 y = 0