integral of (3x-2)(x+3)

Lösung

\int (3 x - 2) (x + 3) d x

x^{3} + \frac{7 x ^{2}}{2} - 6 x + C

Schritte zur Lösung

\int (3 x - 2) (x + 3) d x

Multipliziere aus

(3 x - 2) (x + 3) : 3 x^{2} + 7 x - 6

= \int 3 x^{2} + 7 x - 6 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 3 x^{2} d x + \int 7 x d x - \int 6 d x

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

\int 7 x d x = \frac{7 x ^{2}}{2}

\int 6 d x = 6 x

= x^{3} + \frac{7 x ^{2}}{2} - 6 x

Füge eine Konstante zur Lösung hinzu

= x^{3} + \frac{7 x ^{2}}{2} - 6 x + C

\int \frac{x ^{4}}{x ^{5} - 8} d x

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

\int \frac{4 + 6 x}{1 + x ^{2}} d x

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

\int \frac{x ^{2} - 8 x + 1}{x ^{2} + 1} d x