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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

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

Take the constant out:

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

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

Use the nonelementary integral:

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

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

Simplify

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

Compute the boundaries:

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

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

\int_{2}^{4} e^{x^{2}} d x

\int_{0}^{1} \frac{1}{( x + 1 ) ^{5}} d x

\int_{2}^{3} \frac{x ^{2}}{x ^{2} - 1} d x

\int_{0}^{4} x^{2} - 9 d x

\int_{0}^{1} 4 x^{2} + 1 x d x

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