integral of (e^t+1)^2

Lösung

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

\frac{1}{2} e^{2 t} + 2 e^{t} + t + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Multipliziere aus

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

= \int u + 2 + \frac{1}{u} 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 2 d u + \int \frac{1}{u} d u

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} + 2 u + ln ∣ u ∣

Setze in

u = e^{t}

ein

= \frac{( e ^{t} ) ^{2}}{2} + 2 e^{t} + ln e^{t}

Vereinfache

\frac{( e ^{t} ) ^{2}}{2} + 2 e^{t} + ln e^{t} : \frac{1}{2} e^{2 t} + 2 e^{t} + t

= \frac{1}{2} e^{2 t} + 2 e^{t} + t

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} e^{2 t} + 2 e^{t} + t + C

x \to 2 lim (\frac{1}{x + 1})

in v erse l a pl a ce \frac{3 s}{s ^{2} + 3 s + 16}

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