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

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

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integral from 0 to 1 of (e^{-x})/x

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

- Ei (0) + Ei (- 1)

Solution steps

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

Apply u-substitution

: \int_{0}^{- 1} \frac{e ^{u}}{u} d u

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

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

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

Use the nonelementary integral:

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

= - [Ei (u)]_{- 1}^{0}

Compute the boundaries:

Ei (0) - Ei (- 1)

= - (Ei (0) - Ei (- 1))

Simplify

= - Ei (0) + Ei (- 1)

\frac{d y}{d x} = \frac{y ^{2} x}{x ^{2} + 1}

6 x - 2 x^{2} = y, (3, 4)

y^{^{''}} - 8 y^{^{'}} + 16 y = \frac{12 e ^{4 t}}{( t ^{2} + 1 )}

y = sin (x) cos (x), at x = \frac{3 π}{4}

θ \to \frac{π}{2} lim (3 sin (θ))

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