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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{1/x}x^{-2}

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

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

Solution steps

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

Apply u-substitution

: \int - e^{u} d u

= \int - e^{u} d u

Take the constant out:

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

= - \int e^{u} d u

Use the common integral:

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

= - e^{u}

Substitute back

u = \frac{1}{x}

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

Add a constant to the solution

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

\int 18 x^{3} d x

\int \frac{x ^{3} - 4 x ^{2} + 3}{x - 3} d x

\int (\frac{x ^{3} + 4 x ^{2} + 8 x + 5}{x ^{2} + 2 x + 1}) d x

\int \frac{1}{x ^{2} - 4} d x

\int x (3 x^{2} - 1) d x

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