What is the integral from 0 to 1 of (e^x)/(1+x) ?

The integral from 0 to 1 of (e^x)/(1+x) is 1/e (\Ei(2)-\Ei(1))

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integral from 0 to 1 of (e^x)/(1+x)

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

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

Solution steps

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

Apply u-substitution

: \int_{1}^{2} \frac{e ^{u - 1}}{u} d u

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

Simplify

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

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Use the nonelementary integral:

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

= \frac{1}{e} [Ei (u)]_{1}^{2}

Compute the boundaries:

Ei (2) - Ei (1)

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

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