What is the integral from 0 to 2 of 1/(ln(x)) ?

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

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integral from 0 to 2 of 1/(ln(x))

\int_{0}^{2} \frac{1}{ln ( x )} d x

Ei (ln (1)) + Ei (ln (2)) - Ei (0)

Solution steps

\int_{0}^{2} \frac{1}{ln ( x )} d x

Find undefined (singularity) points:

x = 1

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{1} \frac{1}{ln ( x )} d x + \int_{1}^{2} \frac{1}{ln ( x )} d x

\int_{0}^{1} \frac{1}{ln ( x )} d x = Ei (ln (1))

\int_{1}^{2} \frac{1}{ln ( x )} d x = Ei (ln (2)) - Ei (0)

= Ei (ln (1)) + Ei (ln (2)) - Ei (0)

\int_{0}^{l n (3)} 6 e^{3 x} d x

\int_{e}^{e^{2}} x ln (x) d x

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

\int_{- 1}^{0} x^{2} (4 x^{3} + 5)^{3} d x

\int_{1}^{3} 2 x - 1 d x

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