What is the integral from 2 to 3 of 1/(ln(x-1)) ?

The integral from 2 to 3 of 1/(ln(x-1)) is \Ei(ln(2))-\Ei(0)

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integral from 2 to 3 of 1/(ln(x-1))

\int_{2}^{3} \frac{1}{ln ( x - 1 )} d x

Ei (ln (2)) - Ei (0)

Solution steps

\int_{2}^{3} \frac{1}{ln ( x - 1 )} d x

Apply u-substitution

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

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

Apply u-substitution

: \int_{0}^{l n (2)} \frac{e ^{v}}{v} d v

= \int_{0}^{l n (2)} \frac{e ^{v}}{v} d v

Use the nonelementary integral:

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

= [Ei (v)]_{0}^{l n (2)}

Compute the boundaries:

Ei (ln (2)) - Ei (0)

= Ei (ln (2)) - Ei (0)

\int_{4}^{9} \frac{1}{2} x - x d x

\int_{0}^{1} 4 x - 4 x^{2} d x

\int_{- 3}^{2} e^{u + 3} d u

\int_{2}^{3} 3 x^{2} - \frac{x ^{3}}{4} d x

\int_{0}^{0} sin (x) d x

What is the integral from 2 to 3 of 1/(ln(x-1)) ?
The integral from 2 to 3 of 1/(ln(x-1)) is \Ei(ln(2))-\Ei(0)