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

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

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integral of 1/(2e^x)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{2} (- \frac{1}{u})

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{8} \frac{x}{x ^{2} + 36} d x

\frac{d}{d x} (\frac{x}{1 - x})

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

\frac{e ^{- 5 s}}{s ^{2} - 4 s + 13}

y^{^{''}} = 2 + x