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

The integral from 1 to 2 of 1/(xln(x)) is diverges

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

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

diverges

Solution steps

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

Apply u-substitution

= \int_{0}^{l n (2)} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{0}^{l n (2)}

Compute the boundaries:

diverges

= diverges

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

\int_{0}^{\frac{π}{2}} cos^{2} (θ) d θ

\int_{0}^{\frac{π}{4}} tan^{2} (θ) d θ

\int_{0}^{e - 1} ln (x + 1) d x

\int_{0}^{1} \frac{x}{1 + x} d x

What is the integral from 1 to 2 of 1/(xln(x)) ?
The integral from 1 to 2 of 1/(xln(x)) is diverges