integral of (3t+1)(t-2)

Lösung

\int (3 t + 1) (t - 2) d t

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

Schritte zur Lösung

\int (3 t + 1) (t - 2) d t

Wende die partielle Integration an

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

\int 3 (\frac{t ^{2}}{2} - 2 t) d t = 3 (\frac{t ^{3}}{6} - t^{2})

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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