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

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

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

\int e^{- l n (x)} (1 - \frac{1}{x}) d x

ln ∣ x ∣ + \frac{1}{x} + C

Solution steps

\int e^{- l n (x)} (1 - \frac{1}{x}) d x

Simplify

e^{- l n (x)} (1 - \frac{1}{x}) : \frac{x - 1}{x ^{2}}

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

Expand

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - (- \frac{1}{x})

Simplify

= ln ∣ x ∣ + \frac{1}{x}

Add a constant to the solution

= ln ∣ x ∣ + \frac{1}{x} + C

9 e^{x} + x

\frac{d}{d x} (ln (\frac{x - 1}{x + 1}))

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

x \to - 2 lim (\frac{x ^{2} + 8}{x + 2})

x^{^{''}} + 100 x = 0

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