integral of (4x^3)/(2x+3)

Lösung

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

4 (\frac{x ^{3}}{6} - \frac{3 x ^{2}}{8} + \frac{9}{8} x - \frac{27}{16} ln ∣ 2 x + 3 ∣) + C

Schritte zur Lösung

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Lange Division

\frac{x ^{3}}{2 x + 3} : \frac{x ^{2}}{2} - \frac{3 x}{4} + \frac{9}{8} - \frac{27}{8 ( 2 x + 3 )}

= 4 \cdot \int \frac{x ^{2}}{2} - \frac{3 x}{4} + \frac{9}{8} - \frac{27}{8 ( 2 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

= 4 (\int \frac{x ^{2}}{2} d x - \int \frac{3 x}{4} d x + \int \frac{9}{8} d x - \int \frac{27}{8 ( 2 x + 3 )} d x)

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

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

\int \frac{9}{8} d x = \frac{9}{8} x

\int \frac{27}{8 ( 2 x + 3 )} d x = \frac{27}{16} ln ∣ 2 x + 3 ∣

= 4 (\frac{x ^{3}}{6} - \frac{3 x ^{2}}{8} + \frac{9}{8} x - \frac{27}{16} ln ∣ 2 x + 3 ∣)

Füge eine Konstante zur Lösung hinzu

= 4 (\frac{x ^{3}}{6} - \frac{3 x ^{2}}{8} + \frac{9}{8} x - \frac{27}{16} ln ∣ 2 x + 3 ∣) + C

d er i v a t i v e π r^{2} h

\int \frac{1}{x ^{2} x ^{2} - 9} d x

\frac{\partial}{\partial x} (4 x - 2 y + 3)

n = 1 \sum \infty \frac{x ^{n}}{n + 2}

\frac{d}{d x} ((9 x^{2} - 15 x) e^{x})