What is the integral of e^{1/t} ?

The integral of e^{1/t} is e^{1/t}t-\Ei(1/t)+C

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integral of e^{1/t}

\int e^{\frac{1}{t}} d t

e^{\frac{1}{t}} t - Ei (\frac{1}{t}) + C

Solution steps

\int e^{\frac{1}{t}} d t

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = \frac{1}{t}

= - (- \frac{e ^{\frac{1}{t}}}{\frac{1}{t}} - (- Ei (\frac{1}{t})))

Simplify

- (- \frac{e ^{\frac{1}{t}}}{\frac{1}{t}} - (- Ei (\frac{1}{t}))) : e^{\frac{1}{t}} t - Ei (\frac{1}{t})

= e^{\frac{1}{t}} t - Ei (\frac{1}{t})

Add a constant to the solution

= e^{\frac{1}{t}} t - Ei (\frac{1}{t}) + C

\frac{d}{d x} ((x^{2} + 3 x - 2)^{4})

x \to 0 lim (\frac{tan ( 6 x )}{sin ( 5 x )})

\frac{\partial}{\partial x} (x^{2} + 16 x w + 64 w^{2})

x \to 3 - lim (\frac{2 x + 1}{x - 3})

x \to 0 - lim (\frac{4}{5 x ^{\frac{1}{3}}})

What is the integral of e^{1/t} ?
The integral of e^{1/t} is e^{1/t}t-\Ei(1/t)+C