integral of t(t^3-1)

Lösung

\int t (t^{3} - 1) d t

\frac{t ^{5}}{5} - \frac{t ^{2}}{2} + C

Schritte zur Lösung

\int t (t^{3} - 1) d t

Multipliziere aus

t (t^{3} - 1) : t^{4} - t

= \int t^{4} - 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^{4} d t - \int t d t

\int t^{4} d t = \frac{t ^{5}}{5}

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

= \frac{t ^{5}}{5} - \frac{t ^{2}}{2}

Füge eine Konstante zur Lösung hinzu

= \frac{t ^{5}}{5} - \frac{t ^{2}}{2} + C

\frac{d}{d x} (\frac{e ^{x} + cos ( x )}{x ^{2}})

3 y^{^{'}} = 2 y

\int \frac{x ^{2}}{25 + x ^{2}} d x

\frac{d}{d x} (7 x^{4} - 2 x^{3} + 8 x + 5)

x \to 0 lim (e^{2 x} - 1)