integral of (x+1)(2x-1)

Lösung

\int (x + 1) (2 x - 1) d x

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

Schritte zur Lösung

\int (x + 1) (2 x - 1) d x

Multipliziere aus

(x + 1) (2 x - 1) : 2 x^{2} + x - 1

= \int 2 x^{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 2 x^{2} d x + \int x d x - \int 1 d x

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

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

\int 1 d x = x

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

Füge eine Konstante zur Lösung hinzu

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

d er i v a t i v e \frac{1}{x + 6}

\int \frac{1}{x ( ln ( x ) ) ^{15}} d x

\frac{d}{d x} ((2 x^{3} - 7)^{8})

\frac{d}{d x} (ln (\frac{x}{2}) (x - 2))

x \to 0 - lim (\frac{x + 10}{x ^{2}})