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

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

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

integral

\frac{1}{x ^{3} \cdot ( e ^{- \frac{1}{x}} )}

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

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{x ^{3} e ^{- \frac{1}{x}}} = x^{- 3} \frac{1}{e ^{- \frac{1}{x}}}

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

Simplify

x^{- 3} \frac{1}{e ^{- \frac{1}{x}}} : \frac{e ^{\frac{1}{x}}}{x ^{3}}

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

Apply u-substitution

: \int - e^{u} u d u

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

Take the constant out:

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

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

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= - (e^{u} u - \int e^{u} d u)

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

= - (e^{u} u - e^{u})

Substitute back

u = \frac{1}{x}

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

Simplify

- (e^{\frac{1}{x}} \frac{1}{x} - e^{\frac{1}{x}}) : - \frac{e ^{\frac{1}{x}}}{x} + e^{\frac{1}{x}}

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

Add a constant to the solution

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

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

(a^{2} - x^{2})^{\frac{1}{2}}

x^{2} + sin (x)

x^{3} + x + 1

\frac{x ^{3}}{( x - 1 )}

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