What is the integral of (e^{-x^2})/x ?

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

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integral of (e^{-x^2})/x

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

\frac{1}{2} Ei (- x^{2}) + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{- u}}{2 u} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

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

= \frac{1}{2} \cdot \int \frac{e ^{v}}{v} d v

Use the nonelementary integral:

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

= \frac{1}{2} Ei (v)

Substitute back

= \frac{1}{2} Ei (- x^{2})

Add a constant to the solution

= \frac{1}{2} Ei (- x^{2}) + C

\frac{d}{d x} (ln (\frac{1 - x}{1 + x}))

x \to 0 lim ((1 + 5 x)^{\frac{6}{x}})

\int 5 cos (x^{2}) d x

(4, - 1), (- 1, - 4)

\frac{1}{s ^{2} + s ^{1} + s ^{1} + 1}

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