What is the integral of 2e^{-x}+x-(e^x)/2 ?

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

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integral of 2e^{-x}+x-(e^x)/2

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

- 2 e^{- x} + \frac{x ^{2}}{2} - \frac{1}{2} e^{x} + C

Solution steps

\int 2 e^{- x} + x - \frac{e ^{x}}{2} 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

= \int 2 e^{- x} d x + \int x d x - \int \frac{e ^{x}}{2} d x

\int 2 e^{- x} d x = - 2 e^{- x}

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

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

= - 2 e^{- x} + \frac{x ^{2}}{2} - \frac{1}{2} e^{x}

Add a constant to the solution

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

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