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

Take the constant out:

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

= - \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}})

Simplify

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

Add a constant to the solution

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

\int (x - y) d x

\frac{\partial}{\partial x} (e^{4 e^{x}})

x + \frac{2}{x}

\frac{d}{d x} (3^{s i n (x)} + \frac{e ^{- x}}{2})

5 (sin (x^{2}) + 2 x^{2} cos (x^{2}))

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