integral from 1/2 to 2 of 1/(x(x+1))

Lösung

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

ln (2)

Dezimale

0.69314 \dots

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{x ( x + 1 )} : \frac{1}{x} - \frac{1}{x + 1}

= \int_{\frac{1}{2}}^{2} \frac{1}{x} - \frac{1}{x + 1} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{\frac{1}{2}}^{2} \frac{1}{x} d x - \int_{\frac{1}{2}}^{2} \frac{1}{x + 1} d x

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

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

= 2 ln (2) - ln (2)

Vereinfache

= ln (2)

\int_{0}^{5} (x - 5) (x - 3) d x

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\int_{0}^{\frac{π}{2}} \frac{1}{2} d x

\int_{1}^{4} 7 x + x d x