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

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

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

\int e^{- x} + \frac{ln ( x )}{x} d x

- e^{- x} + \frac{ln ^{2} ( x )}{2} + C

Solution steps

\int e^{- x} + \frac{ln ( x )}{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 e^{- x} d x + \int \frac{ln ( x )}{x} d x

\int e^{- x} d x = - e^{- x}

\int \frac{ln ( x )}{x} d x = \frac{ln ^{2} ( x )}{2}

= - e^{- x} + \frac{ln ^{2} ( x )}{2}

Add a constant to the solution

= - e^{- x} + \frac{ln ^{2} ( x )}{2} + C

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