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

Lösung

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

arctan (e) - \frac{π}{4}

Dezimale

0.43288 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Nutze das gemeinsame Integral :

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= [arctan (u)]_{1}^{e}

Berechne die Grenzen:

arctan (e) - \frac{π}{4}

= arctan (e) - \frac{π}{4}

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