integral of (4x-3)(2x-1)

Lösung

\int (4 x - 3) (2 x - 1) d x

\frac{8 x ^{3}}{3} - 5 x^{2} + 3 x + C

Schritte zur Lösung

\int (4 x - 3) (2 x - 1) d x

Multipliziere aus

(4 x - 3) (2 x - 1) : 8 x^{2} - 10 x + 3

= \int 8 x^{2} - 10 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

= \int 8 x^{2} d x - \int 10 x d x + \int 3 d x

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

\int 10 x d x = 5 x^{2}

\int 3 d x = 3 x

= \frac{8 x ^{3}}{3} - 5 x^{2} + 3 x

Füge eine Konstante zur Lösung hinzu

= \frac{8 x ^{3}}{3} - 5 x^{2} + 3 x + C

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

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

d er i v a t i v e f (x) = 8 x^{3}

\frac{d}{d x} ((sec^{2} (x)) d x)

d er i v a t i v e f (t) = t^{6} sin (t)