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

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

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

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

ln ∣ x ∣ - \frac{1}{x} + C

Solution steps

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

Simplify

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}

Add a constant to the solution

= ln ∣ x ∣ - \frac{1}{x} + C

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