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

Lösung

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

2 ln (2) - ln (3)

Dezimale

0.28768 \dots

Schritte zur Lösung

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

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

\int_{1}^{2} \frac{1}{x + 1} d x = ln (3) - ln (2)

= ln (2) - (ln (3) - ln (2))

Vereinfache

= 2 ln (2) - ln (3)

\int_{- 1}^{1} 2 cos (\frac{π}{3} x) d x

\int_{0}^{2} (x^{3}) d x

\int_{0}^{60} 100 e^{- 0.01 t} d t

\int_{1}^{x^{2}} 3 + t^{2} d t

\int_{0}^{1} 2 π x d x