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

Lösung

\int_{0}^{1} \frac{e ^{x} - 1}{e ^{x} + 1} d x

2 (ln (e + 1) - ln (2)) - 1

Dezimale

0.24022 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{e ^{x} - 1}{e ^{x} + 1} d x

\frac{e ^{x} - 1}{e ^{x} + 1} = \frac{2 e ^{x}}{e ^{x} + 1} - 1

= \int_{0}^{1} \frac{2 e ^{x}}{e ^{x} + 1} - 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}^{1} \frac{2 e ^{x}}{e ^{x} + 1} d x - \int_{0}^{1} 1 d x

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

\int_{0}^{1} 1 d x = 1

= 2 (ln (e + 1) - ln (2)) - 1

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