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

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

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

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

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

Solution steps

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

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

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

Take the constant out:

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

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

Simplify

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

Simplify

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

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

Apply u-substitution

: \int e^{u} d u

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

Use the common integral:

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

= e e^{u}

Substitute back

u = - \frac{1}{x}

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add a constant to the solution

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

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