What is the integral of 1/6 (x-1) ?

The integral of 1/6 (x-1) is 1/6 ((x^2)/2-x)+C

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integral of 1/6 (x-1)

\int \frac{1}{6} (x - 1) d x

\frac{1}{6} (\frac{x ^{2}}{2} - x) + C

Solution steps

\int \frac{1}{6} (x - 1) d x

Take the constant out:

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

= \frac{1}{6} \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}{6} (\int x d x - \int 1 d x)

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

\int 1 d x = x

= \frac{1}{6} (\frac{x ^{2}}{2} - x)

Add a constant to the solution

= \frac{1}{6} (\frac{x ^{2}}{2} - x) + C

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What is the integral of 1/6 (x-1) ?
The integral of 1/6 (x-1) is 1/6 ((x^2)/2-x)+C