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

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

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

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

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

Solution steps

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

3 (sin (t) y^{^{'}} + cos (t) y) = cos (t) sin^{2} (t)

\int arctan (9 x) d x

f (x) = \frac{1}{x} + \frac{1}{x ^{2}}

\frac{( 5 x ^{5} - 3 x ^{2} + 1 )}{x}

\int \frac{1}{12 ( x ^{2} + 1 )} d x

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