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

Lösung

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

- 1

Schritte zur Lösung

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Wende U-Substitution an

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

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= - [ln ∣ u ∣]_{\frac{1}{e} + 1}^{e + 1}

Berechne die Grenzen:

ln (e + 1) - ln (\frac{1}{e} + 1)

= - (ln (e + 1) - ln (\frac{1}{e} + 1))

Vereinfache

= - 1

\int_{0}^{2} (x - 1) e^{x} d x

\int_{- 1}^{1} x^{6} sin (x) cos (x) d x

\int_{e^{9}}^{e^{49}} \frac{1}{x ln ( x )} d x

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

\int_{- 6}^{3} - x^{2} - 3 x d x