What is the integral of (e^{-1/(x^2)})/(x^3) ?

The integral of (e^{-1/(x^2)})/(x^3) is 1/2 e^{-1/(x^2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (e^{-1/(x^2)})/(x^3)

\int \frac{e ^{- \frac{1}{x ^{2}}}}{x ^{3}} d x

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

Solution steps

\int \frac{e ^{- \frac{1}{x ^{2}}}}{x ^{3}} d x

Apply u-substitution

= \int \frac{e ^{u}}{2} 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 = - \frac{1}{x ^{2}}

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

Add a constant to the solution

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

What is the integral of (e^{-1/(x^2)})/(x^3) ?
The integral of (e^{-1/(x^2)})/(x^3) is 1/2 e^{-1/(x^2)}+C