What is the integral from 2 to x of 1/(ln(x)) ?

The integral from 2 to x of 1/(ln(x)) is \Ei(ln(x))-\Ei(ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 2 to x of 1/(ln(x))

\int_{2}^{x} \frac{1}{ln ( x )} d x

Ei (ln (x)) - Ei (ln (2))

Solution steps

\int_{2}^{x} \frac{1}{ln ( x )} d x

Apply u-substitution

: \int_{l n (2)}^{l n (x)} \frac{e ^{u}}{u} d u

= \int_{l n (2)}^{l n (x)} \frac{e ^{u}}{u} d u

Use the nonelementary integral:

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

= [Ei (u)]_{l n (2)}^{l n (x)}

Compute the boundaries:

Ei (ln (x)) - Ei (ln (2))

= Ei (ln (x)) - Ei (ln (2))

\int_{- 1}^{1} 2 x^{4} d x

\int_{0}^{\frac{π}{3}} (3 sec^{2} (x)) d x

\int_{1}^{2} \frac{x}{4} d x

\int_{0}^{\frac{π}{2}} 8 cos (x) d x

\int_{- 3}^{3} ∣ x + 1 ∣ d x

What is the integral from 2 to x of 1/(ln(x)) ?
The integral from 2 to x of 1/(ln(x)) is \Ei(ln(x))-\Ei(ln(2))