integral from 0 to ln(2) of (e^{3x})/(1+e^{6x)}

Lösung

\int_{0}^{l n (2)} \frac{e ^{3 x}}{1 + e ^{6 x}} d x

\frac{1}{3} (arctan (8) - \frac{π}{4})

Dezimale

0.22034 \dots

Schritte zur Lösung

\int_{0}^{l n (2)} \frac{e ^{3 x}}{1 + e ^{6 x}} d x

Wende U-Substitution an

= \int_{2}^{65} \frac{1}{6 u u - 1} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int_{2}^{65} \frac{1}{u u - 1} d u

Wende U-Substitution an

= \frac{1}{6} \cdot \int_{1}^{8} \frac{2}{v ^{2} + 1} d v

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot 2 \cdot \int_{1}^{8} \frac{1}{v ^{2} + 1} d v

Nutze das gemeinsame Integral :

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

= \frac{1}{6} \cdot 2 [arctan (v)]_{1}^{8}

Vereinfache

\frac{1}{6} \cdot 2 [arctan (v)]_{1}^{8} : \frac{1}{3} [arctan (v)]_{1}^{8}

= \frac{1}{3} [arctan (v)]_{1}^{8}

Berechne die Grenzen:

arctan (8) - \frac{π}{4}

= \frac{1}{3} (arctan (8) - \frac{π}{4})

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