What is the integral from 0 to 1 of x^{-1}ln(x) ?

The integral from 0 to 1 of x^{-1}ln(x) is diverges

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integral from 0 to 1 of x^{-1}ln(x)

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

diverges

Solution steps

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

Simplify

x^{- 1} ln (x) : \frac{ln ( x )}{x}

= \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

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

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\int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{2} (x) d x

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

What is the integral from 0 to 1 of x^{-1}ln(x) ?
The integral from 0 to 1 of x^{-1}ln(x) is diverges