integral of (x^3-6x-4)/(x+2)

Lösung

\int \frac{x ^{3} - 6 x - 4}{x + 2} d x

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

Schritte zur Lösung

\int \frac{x ^{3} - 6 x - 4}{x + 2} d x

Vereinfache

\frac{x ^{3} - 6 x - 4}{x + 2} : x^{2} - 2 x - 2

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

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

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

\int 2 d x = 2 x

= \frac{x ^{3}}{3} - x^{2} - 2 x

Füge eine Konstante zur Lösung hinzu

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

\int \frac{4 y ^{2} - 6 y - 12}{y ( y + 2 ) ( y - 3 )} d y

\int e^{\frac{x}{x}} d x

\int - \frac{1}{2 ( x + 1 )} d x

\int (4 - 5 x) d x

\int d (\frac{x}{y})