integral from 1 to 2 of (ln(t))/(t^2)

Lösung

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

\frac{- ln ( 2 ) - 1}{2} + 1

Dezimale

0.15342 \dots

Schritte zur Lösung

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

Wende die partielle Integration an

= [- \frac{ln ( t )}{t} - \int - \frac{1}{t ^{2}} d t]_{1}^{2}

\int - \frac{1}{t ^{2}} d t = \frac{1}{t}

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

Da die Nenner gleich sind, fasse die Brüche zusammen.:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Berechne die Grenzen:

\frac{- ln ( 2 ) - 1}{2} + 1

= \frac{- ln ( 2 ) - 1}{2} + 1

\int \frac{sin ( x )}{1 - sin ( x )} d x

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