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

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

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integral of 1/(2e^u)

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

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

Solution steps

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

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 ^{u}} d u

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

v = e^{u}

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

Simplify

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

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

Add a constant to the solution

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

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