integral of sqrt((e^t-1)^2+(2e^{t/2))^2}

Lösung

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

\frac{2}{3} e^{t} + 2 + \frac{1}{3} e^{t} + t + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende die partielle Integration an

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

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

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

Setze in

u = e^{\frac{t}{2}}

ein

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

Vereinfache

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

= \frac{2}{3} e^{t} + 2 + \frac{1}{3} e^{t} + t

Füge eine Konstante zur Lösung hinzu

= \frac{2}{3} e^{t} + 2 + \frac{1}{3} e^{t} + t + C

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