integral of (8+t+t^2)/(sqrt(t))

Lösung

\int \frac{8 + t + t ^{2}}{t} d t

16 t + \frac{2}{3} t^{\frac{3}{2}} + \frac{2}{5} t^{\frac{5}{2}} + C

Schritte zur Lösung

\int \frac{8 + t + t ^{2}}{t} d t

Multipliziere aus

\frac{8 + t + t ^{2}}{t} : \frac{8}{t} + t + t^{\frac{3}{2}}

= \int \frac{8}{t} + t + t^{\frac{3}{2}} 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 \frac{8}{t} d t + \int t d t + \int t^{\frac{3}{2}} d t

\int \frac{8}{t} d t = 16 t

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

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

= 16 t + \frac{2}{3} t^{\frac{3}{2}} + \frac{2}{5} t^{\frac{5}{2}}

Füge eine Konstante zur Lösung hinzu

= 16 t + \frac{2}{3} t^{\frac{3}{2}} + \frac{2}{5} t^{\frac{5}{2}} + C

t an g e n t 8 x^{2} + 3 x

d er i v a t i v e f (x) = 2 x^{2} - x^{3}

d er i v a t i v e (2 t - 1) (4 t - 6)^{- 1}

x^{2} y^{^{''}} - 2 y = 0, y (1) = 0, y^{^{'}} (1) = 3

x \to + 0 lim (x^{8} - \frac{1}{x})