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

The integral of 1/(2e^{2x)} is -1/4 e^{-2x}+C

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

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

- \frac{1}{4} e^{- 2 x} + C

Solution steps

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

Apply exponent rule:

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

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

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = - 2 x

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

Simplify

\frac{1}{2} (- \frac{1}{2} e^{- 2 x}) : - \frac{1}{4} e^{- 2 x}

= - \frac{1}{4} e^{- 2 x}

Add a constant to the solution

= - \frac{1}{4} e^{- 2 x} + C

\int_{0}^{1} π (x - x^{3})^{2} d x

2 x^{3} - 6 x^{2} - 18 - 3

y = - 2 x^{2} + 6, y = - 2

\int 1 + x^{2} \cdot x^{5} d x

What is the integral of 1/(2e^{2x)} ?
The integral of 1/(2e^{2x)} is -1/4 e^{-2x}+C