integral of 1/(x^3+3x^2)

Lösung

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

- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{x ^{3} + 3 x ^{2}} : - \frac{1}{9 x} + \frac{1}{3 x ^{2}} + \frac{1}{9 ( x + 3 )}

= \int - \frac{1}{9 x} + \frac{1}{3 x ^{2}} + \frac{1}{9 ( x + 3 )} 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 \frac{1}{9 x} d x + \int \frac{1}{3 x ^{2}} d x + \int \frac{1}{9 ( x + 3 )} d x

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

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

\int \frac{1}{9 ( x + 3 )} d x = \frac{1}{9} ln ∣ x + 3 ∣

= - \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣ + C

y^{^{''}} - y^{^{'}} - 2 y = - 6 t + 10 t^{2}

in v erse l a pl a ce \frac{1 + e ^{- s π}}{s ^{2} + 1}

d er i v a t i v e y = 3 e^{3 e^{x} + x}

a re a tan (x), 2 sin (x), [- \frac{π}{3}, \frac{π}{3}]

\frac{\partial}{\partial x} (e^{- 2 x})