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

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

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

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

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

Solution steps

\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}

= \int e^{- 2 x} d x

Apply u-substitution

= \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 e^{u} d u

Use the common integral:

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

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

Substitute back

u = - 2 x

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (- 3 y cos (- 9 x))

\frac{d}{d x} (\frac{2}{4 x + 1})

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

d er i v a t i v e 4 x e^{- k x}

\frac{\partial}{\partial x} (2 x y z^{2} - 192)

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