integral of (\sqrt[3]{x}+1)^2

Lösung

\int (3 x + 1)^{2} d x

\frac{3}{5} x^{\frac{5}{3}} + \frac{3}{2} x^{\frac{4}{3}} + x + C

Schritte zur Lösung

\int (3 x + 1)^{2} d x

Multipliziere aus

(3 x + 1)^{2} : x^{\frac{2}{3}} + 2 3 x + 1

= \int x^{\frac{2}{3}} + 2 3 x + 1 d x

Wende die Summenregel an:

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

= \int x^{\frac{2}{3}} d x + \int 2 3 x d x + \int 1 d x

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

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

\int 1 d x = x

= \frac{3}{5} x^{\frac{5}{3}} + \frac{3}{2} x^{\frac{4}{3}} + x

Füge eine Konstante zur Lösung hinzu

= \frac{3}{5} x^{\frac{5}{3}} + \frac{3}{2} x^{\frac{4}{3}} + x + C

t^{2} y^{^{'}} + t y = 3, y (1) = 6

d er i v a t i v e f (x) = 9 x

a re a y = - x, y = \frac{1}{2} x - \frac{3}{2}, x = - 1, x = 3

d er i v a t i v e \frac{x ^{2}}{( 3 + x )}

\int \frac{1}{9 x + 4} d x