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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to x^2 of 1/(1+ln(t))

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

\frac{1}{e} (Ei (1 + 2 ln (x)) - Ei (1))

Solution steps

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

Apply u-substitution

: \int_{1}^{1 + l n (x^{2})} \frac{e ^{u - 1}}{u} d u

= \int_{1}^{1 + l n (x^{2})} \frac{e ^{u - 1}}{u} d u

Simplify

e^{u - 1} : e^{u} e^{- 1}

= \int_{1}^{1 + l n (x^{2})} \frac{e ^{u} e ^{- 1}}{u} d u

Take the constant out:

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

= e^{- 1} \cdot \int_{1}^{1 + l n (x^{2})} \frac{e ^{u}}{u} d u

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{e} \cdot \int_{1}^{1 + l n (x^{2})} \frac{e ^{u}}{u} d u

Use the nonelementary integral:

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

= \frac{1}{e} [Ei (u)]_{1}^{1 + l n (x^{2})}

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= \frac{1}{e} [Ei (u)]_{1}^{1 + 2 l n (x)}

Compute the boundaries:

Ei (1 + 2 ln (x)) - Ei (1)

= \frac{1}{e} (Ei (1 + 2 ln (x)) - Ei (1))

\int_{0}^{1} 1 - 2 x d x

\int_{- π}^{0} x \cdot sin (n x) d x

\int_{1}^{x} cos (t) - 1 d t

\int_{2}^{4} x^{2} + x + 2 d x

\int_{0}^{5} r e^{- r^{2}} d r

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