integral of t(2t-3sqrt(t))

Lösung

\int t (2 t - 3 t) d t

2 (\frac{1}{5} t^{\frac{5}{2}} (2 t - 3) - \frac{t ^{3}}{15}) + C

Schritte zur Lösung

\int t (2 t - 3 t) d t

Wende U-Substitution an

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

Entferne die Konstante:

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

= 2 \cdot \int u^{3} (2 u^{2} - 3 u) d u

Multipliziere mit der Konjugation von

2 u^{2} - 3 u : \frac{4 u ^{4} - 9 u ^{2}}{2 u ^{2} + 3 u}

= 2 \cdot \int u^{3} \frac{4 u ^{4} - 9 u ^{2}}{2 u ^{2} + 3 u} d u

Vereinfache

u^{3} \frac{4 u ^{4} - 9 u ^{2}}{2 u ^{2} + 3 u} : u^{4} (2 u - 3)

= 2 \cdot \int u^{4} (2 u - 3) d u

Wende die partielle Integration an

= 2 (\frac{1}{5} u^{5} (2 u - 3) - \int \frac{2 u ^{5}}{5} d u)

\int \frac{2 u ^{5}}{5} d u = \frac{u ^{6}}{15}

= 2 (\frac{1}{5} u^{5} (2 u - 3) - \frac{u ^{6}}{15})

Setze in

u = t

ein

= 2 (\frac{1}{5} (t)^{5} (2 t - 3) - \frac{( t ) ^{6}}{15})

Vereinfache

2 (\frac{1}{5} (t)^{5} (2 t - 3) - \frac{( t ) ^{6}}{15}) : 2 (\frac{1}{5} t^{\frac{5}{2}} (2 t - 3) - \frac{t ^{3}}{15})

= 2 (\frac{1}{5} t^{\frac{5}{2}} (2 t - 3) - \frac{t ^{3}}{15})

Füge eine Konstante zur Lösung hinzu

= 2 (\frac{1}{5} t^{\frac{5}{2}} (2 t - 3) - \frac{t ^{3}}{15}) + C

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