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

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

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

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

ln ∣ x ∣ - \frac{1}{2} e^{- 2 x} + C

Solution steps

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

\int x^{- 1} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - \frac{1}{2} e^{- 2 x}

Add a constant to the solution

= ln ∣ x ∣ - \frac{1}{2} e^{- 2 x} + C

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