integral of (x+4)(2x+1)

Lösung

\int (x + 4) (2 x + 1) d x

\frac{2 x ^{3}}{3} + \frac{9 x ^{2}}{2} + 4 x + C

Schritte zur Lösung

\int (x + 4) (2 x + 1) d x

Multipliziere aus

(x + 4) (2 x + 1) : 2 x^{2} + 9 x + 4

= \int 2 x^{2} + 9 x + 4 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 9 x d x + \int 4 d x

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

\int 9 x d x = \frac{9 x ^{2}}{2}

\int 4 d x = 4 x

= \frac{2 x ^{3}}{3} + \frac{9 x ^{2}}{2} + 4 x

Füge eine Konstante zur Lösung hinzu

= \frac{2 x ^{3}}{3} + \frac{9 x ^{2}}{2} + 4 x + C

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

\frac{d}{d x} (cos (θ) x)

\int 5 x d x

\frac{\partial}{\partial y} (sin^{2} (x y))

y^{^{'}} + \frac{4 y}{x} = - \frac{3}{x}, y (1) = - 2