What is the integral of e^{4/x} ?

The integral of e^{4/x} is e^{4/x}x-4\Ei(4/x)+C

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integral of e^{4/x}

\int e^{\frac{4}{x}} d x

e^{\frac{4}{x}} x - 4 Ei (\frac{4}{x}) + C

Solution steps

\int e^{\frac{4}{x}} d x

Apply u-substitution

: \int - \frac{4 e ^{u}}{u ^{2}} d u

= \int - \frac{4 e ^{u}}{u ^{2}} d u

Take the constant out:

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

= - 4 \cdot \int \frac{e ^{u}}{u ^{2}} d u

Apply Integration By Parts

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

= - 4 (- \frac{e ^{u}}{u} - \int - \frac{e ^{u}}{u} d u)

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

= - 4 (- \frac{e ^{u}}{u} - (- Ei (u)))

Substitute back

u = \frac{4}{x}

= - 4 (- \frac{e ^{\frac{4}{x}}}{\frac{4}{x}} - (- Ei (\frac{4}{x})))

Simplify

- 4 (- \frac{e ^{\frac{4}{x}}}{\frac{4}{x}} - (- Ei (\frac{4}{x}))) : e^{\frac{4}{x}} x - 4 Ei (\frac{4}{x})

= e^{\frac{4}{x}} x - 4 Ei (\frac{4}{x})

Add a constant to the solution

= e^{\frac{4}{x}} x - 4 Ei (\frac{4}{x}) + C

ln (2 x - x^{2})

y = x, y = \frac{1}{x ^{2}}, x = 2

y^{^{'}} + \frac{2 y}{x} = - \frac{4}{x}, y (1) = 3

(\frac{1}{e ^{x}})^{^{'}}

\frac{d}{d x} (x + x + 8)

What is the integral of e^{4/x} ?
The integral of e^{4/x} is e^{4/x}x-4\Ei(4/x)+C