integral of (x^3-3x^2+4x-9)/(x^2+3)

Lösung

\int \frac{x ^{3} - 3 x ^{2} + 4 x - 9}{x ^{2} + 3} d x

\frac{x ^{2}}{2} - 3 x + \frac{1}{2} ln x^{2} + 3 + C

Schritte zur Lösung

\int \frac{x ^{3} - 3 x ^{2} + 4 x - 9}{x ^{2} + 3} d x

Lange Division

\frac{x ^{3} - 3 x ^{2} + 4 x - 9}{x ^{2} + 3} : x - 3 + \frac{x}{x ^{2} + 3}

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

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

\int 3 d x = 3 x

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

= \frac{x ^{2}}{2} - 3 x + \frac{1}{2} ln x^{2} + 3

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{2}}{2} - 3 x + \frac{1}{2} ln x^{2} + 3 + C

d er i v a t i v e - cos (2 x)

\int_{1}^{4} \frac{x - 2}{x} d x

d er i v a t i v e f (x) = 3 arccsc (x)

\int x^{2} sech^{2} (x^{3}) d x

\frac{d}{d x} (x^{3} - 8)