What is the integral of 1/(e^xx^2) ?

The integral of 1/(e^xx^2) is -(e^{-x})/x-\Ei(-x)+C

integral of 1/(e^xx^2)

\int \frac{1}{e ^{x} x ^{2}} d x

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

Solution steps

\int \frac{1}{e ^{x} x ^{2}} d x

Apply exponent rule:

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

\frac{1}{e ^{x} x ^{2}} = e^{- x} \frac{1}{x ^{2}}

= \int e^{- x} \frac{1}{x ^{2}} d x

Simplify

e^{- x} \frac{1}{x ^{2}} : \frac{e ^{- x}}{x ^{2}}

= \int \frac{e ^{- x}}{x ^{2}} d x

Apply Integration By Parts

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

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

= - \frac{e ^{- x}}{x} - Ei (- x)

Add a constant to the solution

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

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What is the integral of 1/(e^xx^2) ?
The integral of 1/(e^xx^2) is -(e^{-x})/x-\Ei(-x)+C