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

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

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integral of ((x-1))/2

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

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

Solution steps

\int \frac{x - 1}{2} d x

Take the constant out:

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

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

(8 + x) y^{^{'}} = 7 y

\frac{\partial}{\partial x} (ln (x^{2} + y^{4}) - z)

\frac{\partial}{\partial x} (\frac{2 y}{9 x ^{2} + y ^{2} + 9})

ln (x^{2} + 9)

\frac{d}{d x} (2 e^{- x^{2}})

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