What is the integral from 0 to 1 of 1/(xe^{x^2)} ?

The integral from 0 to 1 of 1/(xe^{x^2)} is -1/2 (\Ei(0)-\Ei(-1))

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integral from 0 to 1 of 1/(xe^{x^2)}

\int_{0}^{1} \frac{1}{x e ^{x^{2}}} d x

- \frac{1}{2} (Ei (0) - Ei (- 1))

Solution steps

\int_{0}^{1} \frac{1}{x e ^{x^{2}}} d x

Apply u-substitution

= \int_{0}^{1} \frac{1}{2 e ^{u} 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_{0}^{1} \frac{1}{e ^{u} u} d u

Apply exponent rule:

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

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

= \frac{1}{2} \cdot \int_{0}^{1} e^{- u} \frac{1}{u} d u

Simplify

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

= \frac{1}{2} \cdot \int_{0}^{1} \frac{e ^{- u}}{u} d u

Apply u-substitution

= \frac{1}{2} \cdot \int_{0}^{- 1} \frac{e ^{v}}{v} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{2} (- \int_{- 1}^{0} \frac{e ^{v}}{v} d v)

Use the nonelementary integral:

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

= \frac{1}{2} (- [Ei (v)]_{- 1}^{0})

Simplify

= - \frac{1}{2} [Ei (v)]_{- 1}^{0}

Compute the boundaries:

Ei (0) - Ei (- 1)

= - \frac{1}{2} (Ei (0) - Ei (- 1))

What is the integral from 0 to 1 of 1/(xe^{x^2)} ?
The integral from 0 to 1 of 1/(xe^{x^2)} is -1/2 (\Ei(0)-\Ei(-1))