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

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

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integral of (1/x-1/(x^2))

\int (\frac{1}{x} - \frac{1}{x ^{2}}) d x

ln ∣ x ∣ + \frac{1}{x} + C

Solution steps

\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

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