integral from 0 to ln(2) of 1/(1+e^x)

Lösung

\int_{0}^{l n (2)} \frac{1}{1 + e ^{x}} d x

- ln (3) + 2 ln (2)

Dezimale

0.28768 \dots

Schritte zur Lösung

\int_{0}^{l n (2)} \frac{1}{1 + e ^{x}} d x

\frac{1}{1 + e ^{x}} = - \frac{e ^{x}}{1 + e ^{x}} + 1

= \int_{0}^{l n (2)} - \frac{e ^{x}}{1 + e ^{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_{0}^{l n (2)} \frac{e ^{x}}{1 + e ^{x}} d x + \int_{0}^{l n (2)} 1 d x

\int_{0}^{l n (2)} \frac{e ^{x}}{1 + e ^{x}} d x = ln (3) - ln (2)

\int_{0}^{l n (2)} 1 d x = ln (2)

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

Vereinfache

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

\int_{4}^{6} \frac{1}{2 x ^{2} - 16 x + 40} d x

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