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

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

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integral of 1/x e^{-x}

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

Ei (- x) + C

Solution steps

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

Simplify

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

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

Apply u-substitution

: \int \frac{e ^{u}}{u} d u

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

Use the nonelementary integral:

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

= Ei (u)

Substitute back

u = - x

= Ei (- x)

Add a constant to the solution

= Ei (- x) + C

n = 0 \sum \infty cos (5 nπ)

f (x) = ln^{5} (x)

y^{^{'}} - 4 y = 10 sin (2 t)

\int (\frac{2}{y + 1})^{2} d y

f (x) = x^{2} - 8, (- 2, - 4)