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

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

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

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

- e^{\frac{1}{x}} - ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{e ^{\frac{1}{x}} - x}{x ^{2}} : \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{1}{x}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= - e^{\frac{1}{x}} - ln ∣ x ∣

Add a constant to the solution

= - e^{\frac{1}{x}} - ln ∣ x ∣ + C

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