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

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

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integral of 1/2 (e^x-1)

\int \frac{1}{2} (e^{x} - 1) d x

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

Solution steps

\int \frac{1}{2} (e^{x} - 1) 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

x \to 0 lim (\frac{∣ x ∣}{x})

x \to 2 lim (x + 2)

f (x) = \frac{1}{x - 4}

\frac{d ^{3}}{d x ^{3}} (\frac{2}{x ^{2}})

\frac{\partial}{\partial x} (2 x e^{- y})

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