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Beliebt Rechnen >

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

Lösung

\int (e^{4 x} + e^{- 4 x})^{2} d x

Lösung

\frac{1}{4} (\frac{1}{2} e^{8 x} + 8 x - \frac{1}{2 e ^{8 x}}) + C

Schritte zur Lösung

\int (e^{4 x} + e^{- 4 x})^{2} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

\frac{( u ^{2} + 1 ) ^{2}}{u ^{3}} : u + \frac{2}{u} + \frac{1}{u ^{3}}

= \frac{1}{4} \cdot \int u + \frac{2}{u} + \frac{1}{u ^{3}} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int u d u = \frac{u ^{2}}{2}

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

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

Setze in

u = e^{4 x}

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

Graph

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