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

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

= a \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}}

= a \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

= a (\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}}

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

Setze in

u = e^{\frac{x}{a}}

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int 5^{\frac{x}{2}} d x

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

in t e g r a l 2 x^{- 1}

\int \frac{x + 2}{7 - 9 x ^{2}} d x

\int sin^{4} (x) x d x