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

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

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integral of (1+x)/2

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

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

Solution steps

\int \frac{1 + x}{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 1 + x 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 1 d x + \int x d x)

\int 1 d x = x

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

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

Add a constant to the solution

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

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