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Beliebt Rechnen >

integral of 1/(1+e^{x-1)}

Lösung

\int \frac{1}{1 + e ^{x - 1}} d x

Lösung

- ln 1 + e^{x - 1} + x - 1 + C

Schritte zur Lösung

\int \frac{1}{1 + e ^{x - 1}} d x

Wende U-Substitution an

= \int \frac{1}{1 + e ^{u}} d u

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

= \int - \frac{e ^{u}}{1 + e ^{u}} + 1 d u

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{e ^{u}}{1 + e ^{u}} d u + \int 1 d u

\int \frac{e ^{u}}{1 + e ^{u}} d u = ln ∣ 1 + e^{u} ∣

\int 1 d u = u

= - ln ∣ 1 + e^{u} ∣ + u

Setze in

u = x - 1

ein

= - ln 1 + e^{x - 1} + x - 1

Füge eine Konstante zur Lösung hinzu

= - ln 1 + e^{x - 1} + x - 1 + C

Graph

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