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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply u-substitution

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

= \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 y

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

Add a constant to the solution

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

x \frac{d y}{d x} + 2 y = x^{3} - x

\int_{0}^{π} t cos^{2} (t) d t

\int sin (\frac{nπ x}{1}) d x

x \to 1 + lim (x + 1)

x \to 1 lim (\frac{x ^{2} + 1}{x - 2})

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