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

Lösung

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

\frac{x ^{3}}{6} - \frac{x ^{2}}{8} + \frac{1}{8} x - \frac{1}{16} ln ∣ 2 x + 1 ∣ + C

Schritte zur Lösung

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

Lange Division

\frac{x ^{3}}{2 x + 1} : \frac{x ^{2}}{2} - \frac{x}{4} + \frac{1}{8} - \frac{1}{8 ( 2 x + 1 )}

= \int \frac{x ^{2}}{2} - \frac{x}{4} + \frac{1}{8} - \frac{1}{8 ( 2 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 \frac{x ^{2}}{2} d x - \int \frac{x}{4} d x + \int \frac{1}{8} d x - \int \frac{1}{8 ( 2 x + 1 )} d x

\int \frac{x ^{2}}{2} d x = \frac{x ^{3}}{6}

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

\int \frac{1}{8} d x = \frac{1}{8} x

\int \frac{1}{8 ( 2 x + 1 )} d x = \frac{1}{16} ln ∣ 2 x + 1 ∣

= \frac{x ^{3}}{6} - \frac{x ^{2}}{8} + \frac{1}{8} x - \frac{1}{16} ln ∣ 2 x + 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{3}}{6} - \frac{x ^{2}}{8} + \frac{1}{8} x - \frac{1}{16} ln ∣ 2 x + 1 ∣ + C

\frac{d}{d y} (2 e^{x y^{2}})

y^{^{'}} = 3 y + x

d er i v a t i v e f (x) = \frac{x}{1 + x}

\int \frac{csc ^{5} ( ln ( x ) )}{x} d x

\frac{d}{d x} (\frac{e ^{x}}{5 x ^{2} + 8})