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

Lösung

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

ln (e^{2} + e) - ln (2 e)

Dezimale

0.62011 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{2 e}^{e^{2} + e} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2 e}^{e^{2} + e}

Berechne die Grenzen:

ln (e^{2} + e) - ln (2 e)

= ln (e^{2} + e) - ln (2 e)

\int_{0}^{2} x 1 + x^{2} d x

\int_{1}^{2} \frac{15 x}{5 x ^{2} + 6} d x

\int_{8}^{\infty} ln (x) d x

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\int_{0}^{3} 2 x + 4 x^{2} + 16 x^{3} d x