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

The integral of 1/(xe^{-x)} is \Ei(x)+C

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

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

Ei (x) + C

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Simplify

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

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

Use the nonelementary integral:

\int \frac{e ^{x}}{x} d x = Ei (x)

= Ei (x)

Add a constant to the solution

= Ei (x) + C

\frac{d}{d x} ((5 x^{3} + 3) (x^{4} - 2 x))

l a pl a ce t r an s f or m t e^{2 t} sin (t)

\frac{d y}{d x} + 2 y = e^{- x}

t a y l or X^{3} - X^{2} + 4 X + 5

t an g e n t f (x) = \frac{125}{x ^{2} + 24}, at x = - 1