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

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

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

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

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

Solution steps

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

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