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

The integral from 0 to 1 of (ln(x))/x is diverges

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

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

diverges

Solution steps

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

Apply Integration By Parts

: [ln (x) - \int \frac{1 - ln ( x )}{x} d x]_{0}^{1}

= [ln (x) - \int \frac{1 - ln ( x )}{x} d x]_{0}^{1}

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

= [ln (x) - (- \frac{( 1 - ln ( x ) ) ^{2}}{2})]_{0}^{1}

Simplify

[ln (x) - (- \frac{( 1 - ln ( x ) ) ^{2}}{2})]_{0}^{1} : [\frac{1}{2} (ln^{2} (x) + 1)]_{0}^{1}

= [\frac{1}{2} (ln^{2} (x) + 1)]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\frac{\partial}{\partial x} (s cos (t))

\int \frac{1}{x ^{4} x ^{2} - 3} d x

\frac{d}{d u} (arctan (\frac{u}{v}))

\frac{d}{d x} (2 cos (4 x))

(x^{2} + 1) \cdot e^{x - 1}, 1

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