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

\frac{\partial}{\partial x} (3 x^{2} + 2 y^{2})

y = x^{2} - 3 x, y = 4 x

(x^{2} + 1) \frac{d y}{d x} + 7 x (y - 1) = 0, y (0) = 9

\int x^{2} e^{21 x} d x

2 y^{^{'}} + t y = 2

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