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

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

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

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

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

Solution steps

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

\int (5 + tan^{2} (x)) d x

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\int 7 x x^{2} + 2 d x

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

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