integral of e^t(1-e^t)^2

Lösung

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

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

Schritte zur Lösung

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

Faktorisiere

1 - e^{t} : - (- 1 + e^{t})

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

(- (- 1 + e^{t}))^{2} = (- 1 + e^{t})^{2}

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

Wende U-Substitution an

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

Multipliziere aus

(- 1 + u)^{2} : 1 - 2 u + u^{2}

= \int 1 - 2 u + u^{2} 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 1 d u - \int 2 u d u + \int u^{2} d u

\int 1 d u = u

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

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

= u - u^{2} + \frac{u ^{3}}{3}

Setze in

u = e^{t}

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int 1 + sin^{2} (x) csc (x) d x

\int \frac{1}{x x + 1} d x

\int sec^{4} (y) tan^{3} (y) d y

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

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