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

\int tan (x) 1 + sec^{4} (x) d x

\int tan^{5} (x) sec^{4} (x) d x

d y + e^{3 y} d x = 0

\frac{d}{d x} (4 sec (x) - 3 x)

d er i v a t i v e 20 ln (3 x + 1) - \frac{x}{3}