integral of t(t^2+1)^2

Lösung

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

\frac{t ^{6}}{6} + \frac{t ^{4}}{2} + \frac{t ^{2}}{2} + C

Schritte zur Lösung

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

Multipliziere aus

t (t^{2} + 1)^{2} : t^{5} + 2 t^{3} + t

= \int t^{5} + 2 t^{3} + t d t

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int t^{5} d t + \int 2 t^{3} d t + \int t d t

\int t^{5} d t = \frac{t ^{6}}{6}

\int 2 t^{3} d t = \frac{t ^{4}}{2}

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

= \frac{t ^{6}}{6} + \frac{t ^{4}}{2} + \frac{t ^{2}}{2}

Füge eine Konstante zur Lösung hinzu

= \frac{t ^{6}}{6} + \frac{t ^{4}}{2} + \frac{t ^{2}}{2} + C

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