What is the integral of (5e^{-1/x})/(x^2) ?

The integral of (5e^{-1/x})/(x^2) is 5e^{-1/x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (5e^{-1/x})/(x^2)

\int \frac{5 e ^{- \frac{1}{x}}}{x ^{2}} d x

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

Solution steps

\int \frac{5 e ^{- \frac{1}{x}}}{x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

Apply u-substitution

= 5 \cdot \int e^{u} d u

Use the common integral:

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

= 5 e^{u}

Substitute back

u = - \frac{1}{x}

= 5 e^{- \frac{1}{x}}

Add a constant to the solution

= 5 e^{- \frac{1}{x}} + C

What is the integral of (5e^{-1/x})/(x^2) ?
The integral of (5e^{-1/x})/(x^2) is 5e^{-1/x}+C