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

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

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integral of (e^x+1)/2

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

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

Solution steps

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\int_{a}^{b} \frac{1}{r} d r

\frac{- x ^{2} - 8}{( x ^{2} - 8 ) ^{2}}

\int_{\frac{1}{3}}^{5} 14 x ln (3 x) d x

\frac{( e ^{- s} )}{s ^{2} - 6 s + 9}

y^{^{'}} = \frac{2 y}{t} + t