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

Lösung

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

divergiert

Schritte zur Lösung

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

Vereinfache

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

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

Wende die partielle Integration an

= [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}

Vereinfache

[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}

Berechne die Grenzen:

divergiert

= divergiert

\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}^{4} 3 d x

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