What is the integral from 0 to 0.1 of 1/(xln|x|) ?

The integral from 0 to 0.1 of 1/(xln|x|) is diverges

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integral from 0 to 0.1 of 1/(xln|x|)

\int_{0}^{0.1} \frac{1}{x ln ∣ x ∣} d x

diverges

Solution steps

\int_{0}^{0.1} \frac{1}{x ln ∣ x ∣} d x

Eliminate Absolutes

= \int_{0}^{0.1} \frac{1}{x ln ( x )} d x

Apply Integration By Parts

: [\frac{1 - ln ( x ) \cdot \int - \frac{l n ( x ) + 1}{x l n ^{2} ( x )} d x}{ln ( x )}]_{0}^{0.1}

= [\frac{1 - ln ( x ) \cdot \int - \frac{l n ( x ) + 1}{x l n ^{2} ( x )} d x}{ln ( x )}]_{0}^{0.1}

\int - \frac{ln ( x ) + 1}{x ln ^{2} ( x )} d x = - ln ∣ ln (x) ∣ + \frac{1}{ln ( x )}

= [\frac{1 - ln ( x ) ( - ln ∣ ln ( x ) ∣ + \frac{1}{l n ( x )} )}{ln ( x )}]_{0}^{0.1}

\frac{1 - ln ( x ) ( - ln ∣ ln ( x ) ∣ + \frac{1}{l n ( x )} )}{ln ( x )} = ln ∣ ln (x) ∣

= [ln ∣ ln (x) ∣]_{0}^{0.1}

Compute the boundaries:

diverges

= diverges

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

\int_{4}^{9} (x + \frac{1}{3 x}) d x

\int_{0}^{1} t^{2} + 1 d t

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

\int_{1}^{2} \frac{t - 1}{t} d t

What is the integral from 0 to 0.1 of 1/(xln|x|) ?
The integral from 0 to 0.1 of 1/(xln|x|) is diverges