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 \frac{e ^{(\frac{1}{x})}}{x ^{2}} d x

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

Solution steps

\int \frac{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 \frac{2 + x}{1 - x} d x

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

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

\int 3 x^{2} + 2 d x

\int \frac{x ^{2} + 6 x + 4}{x ^{4} + 8 x ^{2} + 16} 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