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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

\int 2 d x = 2 x

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d x} (\frac{x}{x + 1})

\int 8 sin^{2} (x - \frac{π}{12}) d x

\frac{d y}{d x} = \frac{( 2 x y - x y ^{2} )}{x ^{2}}

\frac{\partial}{\partial y} (\frac{x y}{x + y})

f (t) = \frac{3}{t}