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

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

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integral of (x-1)^2+1

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

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

Solution steps

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

Expand

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

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

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

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

\int 2 d x = 2 x

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

Add a constant to the solution

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

\frac{( 1 + x ^{2} ) d y}{d x} - (1 + y^{2}) = 0

\int_{0}^{N} e^{- s t} e^{- 5 t} d t

\frac{d}{d x} (\frac{- sin ( x )}{sec ( x ) tan ( x )})

\frac{y d y + x d x}{2 1 + x y} = 0

\frac{\partial}{\partial x} (4 x^{3} y^{\frac{1}{2}})

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