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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int \frac{1}{y} e^{- (y + \frac{x}{y})} d x

- e^{- \frac{x}{y} - y} + C

Solution steps

\int \frac{1}{y} e^{- (y + \frac{x}{y})} d x

Take the constant out:

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

= \frac{1}{y} \cdot \int e^{- (\frac{x}{y} + y)} d x

Simplify

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

Apply u-substitution

: \int e^{- u - y} y d u

= \frac{1}{y} \cdot \int e^{- u - y} y d u

Take the constant out:

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

= \frac{1}{y} y \cdot \int e^{- u - y} d u

Apply u-substitution

: \int - e^{v} d v

= \frac{1}{y} y \cdot \int - e^{v} d v

Take the constant out:

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

= \frac{1}{y} y (- \int e^{v} d v)

Use the common integral:

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

= \frac{1}{y} y (- e^{v})

Substitute back

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

Simplify

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

= - e^{- \frac{x}{y} - y}

Add a constant to the solution

= - e^{- \frac{x}{y} - y} + C

\int - \frac{1}{x ln ( x )} d x

\int e^{x} (sin (x))^{2} d x

\int x^{4} 1 + x^{5} d x

\int cos^{4} (3 t) sin (3 t) d t

\int \frac{y ^{2} + 1}{y ^{2} + 4} d y

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