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

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

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integral of (e^{\frac{-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

Simplify

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

Apply u-substitution

: \int \frac{e ^{u}}{2} d u

= \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

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