integral of ((-2u^4+3sqrt(u)))/(u^2)

Lösung

\int \frac{( - 2 u ^{4} + 3 u )}{u ^{2}} d u

- \frac{2 u ^{3}}{3} - \frac{6}{u} + C

Schritte zur Lösung

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

Multipliziere aus

\frac{- 2 u ^{4} + 3 u}{u ^{2}} : - 2 u^{2} + \frac{3}{u ^{\frac{3}{2}}}

= \int - 2 u^{2} + \frac{3}{u ^{\frac{3}{2}}} d u

Wende die Summenregel an:

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

= - \int 2 u^{2} d u + \int \frac{3}{u ^{\frac{3}{2}}} d u

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

\int \frac{3}{u ^{\frac{3}{2}}} d u = - \frac{6}{u}

= - \frac{2 u ^{3}}{3} - \frac{6}{u}

Füge eine Konstante zur Lösung hinzu

= - \frac{2 u ^{3}}{3} - \frac{6}{u} + C

x \to \frac{1}{4} lim (x^{2} + 3 x)

\frac{d}{d x} (- \frac{3 e ^{\frac{3}{x}}}{x ^{2}})

d er i v a t i v e f (x) = 21 x^{3} ln (x)

\int \frac{2 x ^{4} + 2 5 x}{x} d x

y^{^{''}} + 6 y^{^{'}} + 9 y = 3 t + 5