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

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

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

integral

\frac{( e ^{x} )}{x ^{- 1}}

e^{x} x - e^{x} + C

Solution steps

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

Simplify

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

= \int e^{x} x d x

Apply Integration By Parts

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

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

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

= e^{x} x - e^{x}

Add a constant to the solution

= e^{x} x - e^{x} + C

2 y lo g_{10} (y) + 1

6 x \cdot e^{- 2} x

\frac{t ^{2}}{( 1 + t ^{2} )}

\frac{e ^{4} ln ( x ) - e ^{5} ln ( x )}{x ^{5}}

\frac{1 - x ^{4}}{1 - x}

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