integral of-4x^{3/2}+3x^{2/3}-4x^2

Lösung

\int - 4 x^{\frac{3}{2}} + 3 x^{\frac{2}{3}} - 4 x^{2} d x

- \frac{8}{5} x^{\frac{5}{2}} + \frac{9}{5} x^{\frac{5}{3}} - \frac{4 x ^{3}}{3} + C

Schritte zur Lösung

\int - 4 x^{\frac{3}{2}} + 3 x^{\frac{2}{3}} - 4 x^{2} 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 4 x^{\frac{3}{2}} d x + \int 3 x^{\frac{2}{3}} d x - \int 4 x^{2} d x

\int 4 x^{\frac{3}{2}} d x = \frac{8}{5} x^{\frac{5}{2}}

\int 3 x^{\frac{2}{3}} d x = \frac{9}{5} x^{\frac{5}{3}}

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

= - \frac{8}{5} x^{\frac{5}{2}} + \frac{9}{5} x^{\frac{5}{3}} - \frac{4 x ^{3}}{3}

Füge eine Konstante zur Lösung hinzu

= - \frac{8}{5} x^{\frac{5}{2}} + \frac{9}{5} x^{\frac{5}{3}} - \frac{4 x ^{3}}{3} + C

d er i v a t i v e y = e^{- 6 x^{5}}

\int cot^{6} (x) csc^{4} (x) d x

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

x^{2} y^{^{''}} - 2 x y^{^{'}} + (2 + x^{2}) y = 0

d er i v a t i v e e^{l n (1 + \frac{1}{k})}