What is the integral of 1/(e^xx) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(e^xx)

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

Ei (- x) + C

Solution steps

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

Apply exponent rule:

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

\frac{1}{e ^{x} 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

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

\int \frac{3 x - 1}{x ^{2} + x + 1} d x

\int 6 x^{5} (1 - x^{6})^{2} d x

\int sin (\frac{x}{3}) cos (\frac{x}{3}) d x

\int 20 x^{4} (4 x^{5} + 1)^{6} d x

\int 5 x^{2} + 9 x - 6 d x