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

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

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

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

ln ∣ x ∣ - e^{\frac{1}{x}} + C

Solution steps

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

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

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

Expand

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - e^{\frac{1}{x}}

Add a constant to the solution

= ln ∣ x ∣ - e^{\frac{1}{x}} + C

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