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Beliebt Rechnen >

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

Lösung

\int (e^{x} + \frac{1}{e ^{x}})^{2} d x

Lösung

\frac{1}{2} e^{2 x} + 2 x - \frac{1}{2 e ^{2 x}} + C

Schritte zur Lösung

\int (e^{x} + \frac{1}{e ^{x}})^{2} d x

Wende U-Substitution an

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

Multipliziere aus

\frac{u ^{4} + 2 u ^{2} + 1}{u ^{3}} : u + \frac{2}{u} + \frac{1}{u ^{3}}

= \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

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

Setze in

u = e^{x}

ein

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

Vereinfache

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

= \frac{1}{2} e^{2 x} + 2 x - \frac{1}{2 e ^{2 x}}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} e^{2 x} + 2 x - \frac{1}{2 e ^{2 x}} + C

Graph

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