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Beliebt Rechnen >

integral of (2e^x+e^{-x})/(3e^x-4e^{-x)}

Lösung

\int \frac{2 e ^{x} + e ^{- x}}{3 e ^{x} - 4 e ^{- x}} d x

Lösung

\frac{2}{3} x + \frac{1}{3} ln 2 e^{- x} + 3 + \frac{1}{3} ln 2 e^{- x} - 3 + \frac{1}{8} ln - 4 e^{- 2 x} + 3 + C

Schritte zur Lösung

\int \frac{2 e ^{x} + e ^{- x}}{3 e ^{x} - 4 e ^{- x}} d x

Wende U-Substitution an

= \int - \frac{2 + u ^{2}}{u ( - 4 u ^{2} + 3 )} d u

Entferne die Konstante:

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

= - \int \frac{2 + u ^{2}}{u ( - 4 u ^{2} + 3 )} d u

Multipliziere aus

\frac{2 + u ^{2}}{u ( - 4 u ^{2} + 3 )} : \frac{2}{u ( - 4 u ^{2} + 3 )} + \frac{u ^{2}}{u ( - 4 u ^{2} + 3 )}

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{2}{u ( - 4 u ^{2} + 3 )} d u + \int \frac{u ^{2}}{u ( - 4 u ^{2} + 3 )} d u)

\int \frac{2}{u ( - 4 u ^{2} + 3 )} d u = 2 (\frac{1}{3} ln ∣ u ∣ - \frac{1}{6} ln 2 u + 3 - \frac{1}{6} ln 2 u - 3)

\int \frac{u ^{2}}{u ( - 4 u ^{2} + 3 )} d u = - \frac{1}{8} ln - 4 u^{2} + 3

= - (2 (\frac{1}{3} ln ∣ u ∣ - \frac{1}{6} ln 2 u + 3 - \frac{1}{6} ln 2 u - 3) - \frac{1}{8} ln - 4 u^{2} + 3)

Setze in

u = e^{- x}

ein

= - (2 (\frac{1}{3} ln e^{- x} - \frac{1}{6} ln 2 e^{- x} + 3 - \frac{1}{6} ln 2 e^{- x} - 3) - \frac{1}{8} ln - 4 (e^{- x})^{2} + 3)

Vereinfache

- (2 (\frac{1}{3} ln e^{- x} - \frac{1}{6} ln 2 e^{- x} + 3 - \frac{1}{6} ln 2 e^{- x} - 3) - \frac{1}{8} ln - 4 (e^{- x})^{2} + 3) : \frac{2}{3} x + \frac{1}{3} ln 2 e^{- x} + 3 + \frac{1}{3} ln 2 e^{- x} - 3 + \frac{1}{8} ln - 4 e^{- 2 x} + 3

= \frac{2}{3} x + \frac{1}{3} ln 2 e^{- x} + 3 + \frac{1}{3} ln 2 e^{- x} - 3 + \frac{1}{8} ln - 4 e^{- 2 x} + 3

Füge eine Konstante zur Lösung hinzu

= \frac{2}{3} x + \frac{1}{3} ln 2 e^{- x} + 3 + \frac{1}{3} ln 2 e^{- x} - 3 + \frac{1}{8} ln - 4 e^{- 2 x} + 3 + C

Graph

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