integral of (t^2+\sqrt[4]{t})

Lösung

\int (t^{2} + 4 t) d t

\frac{t ^{3}}{3} + \frac{4}{5} t^{\frac{5}{4}} + C

Schritte zur Lösung

\int t^{2} + 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^{2} d t + \int 4 t d t

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

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

= \frac{t ^{3}}{3} + \frac{4}{5} t^{\frac{5}{4}}

Füge eine Konstante zur Lösung hinzu

= \frac{t ^{3}}{3} + \frac{4}{5} t^{\frac{5}{4}} + C

n = 1 \sum \infty \frac{1}{2 ^{n}}

\frac{\partial}{\partial y} (n sin (2 (m x + n y)))

in v erse l a pl a ce \frac{30}{( 10 s + 1 ) ( s ^{2} + 4 )}

d er i v a t i v e x^{x + 2}

\frac{d}{d x} (\frac{x ^{2} - 64}{x - 8})