integral of (4x+3)(2x^2+3x)

Lösung

\int (4 x + 3) (2 x^{2} + 3 x) d x

2 x^{4} + 6 x^{3} + \frac{9 x ^{2}}{2} + C

Schritte zur Lösung

\int (4 x + 3) (2 x^{2} + 3 x) d x

Multipliziere aus

(4 x + 3) (2 x^{2} + 3 x) : 8 x^{3} + 18 x^{2} + 9 x

= \int 8 x^{3} + 18 x^{2} + 9 x 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^{3} d x + \int 18 x^{2} d x + \int 9 x d x

\int 8 x^{3} d x = 2 x^{4}

\int 18 x^{2} d x = 6 x^{3}

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

= 2 x^{4} + 6 x^{3} + \frac{9 x ^{2}}{2}

Füge eine Konstante zur Lösung hinzu

= 2 x^{4} + 6 x^{3} + \frac{9 x ^{2}}{2} + C

d er i v a t i v e (x + 1)^{2} (x^{2} + 1)^{- 3}

x \to 0 + lim (\frac{x - 2}{∣ x ∣})

d er i v a t i v e T

y^{^{''}} - 2 y + y = \frac{e ^{x}}{1 + x ^{2}}

y^{^{'}} sin (x) + y cos (x) = 0, y (\frac{π}{4}) = 1