What is the integral from 1 to x of (e^{-3t})/t ?

The integral from 1 to x of (e^{-3t})/t is \Ei(-3x)-\Ei(-3)

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integral from 1 to x of (e^{-3t})/t

\int_{1}^{x} \frac{e ^{- 3 t}}{t} d t

Ei (- 3 x) - Ei (- 3)

Solution steps

\int_{1}^{x} \frac{e ^{- 3 t}}{t} d t

Apply u-substitution

: \int_{- 3}^{- 3 x} \frac{e ^{u}}{u} d u

= \int_{- 3}^{- 3 x} \frac{e ^{u}}{u} d u

Use the nonelementary integral:

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

= [Ei (u)]_{- 3}^{- 3 x}

Compute the boundaries:

Ei (- 3 x) - Ei (- 3)

= Ei (- 3 x) - Ei (- 3)

3 x^{2}, (5, 75)

\int \frac{2 - 3 sin ( 2 x )}{cos ( 2 x )} d x

\int \frac{4}{x - 5} d x

\int \frac{x ^{4}}{x ^{3} + 1} d x

x \to - 5 lim (- x + 2)

What is the integral from 1 to x of (e^{-3t})/t ?
The integral from 1 to x of (e^{-3t})/t is \Ei(-3x)-\Ei(-3)