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

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

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integral of 1/(e^{2x+1)}

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

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

Solution steps

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

Simplify

e^{2 x + 1} : e^{2 x} e

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = - 2 x

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

Simplify

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

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

Add a constant to the solution

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

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