What is the integral of (e^x-1-x) ?

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

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integral of (e^x-1-x)

\int (e^{x} - 1 - x) d x

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

Solution steps

\int e^{x} - 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

= \int e^{x} d x - \int 1 d x - \int x d x

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

\int 1 d x = x

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

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

Add a constant to the solution

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

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