integral of 3x^3-5x^2+3x+4

Lösung

\int 3 x^{3} - 5 x^{2} + 3 x + 4 d x

\frac{3 x ^{4}}{4} - \frac{5 x ^{3}}{3} + \frac{3 x ^{2}}{2} + 4 x + C

Schritte zur Lösung

\int 3 x^{3} - 5 x^{2} + 3 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 3 x^{3} d x - \int 5 x^{2} d x + \int 3 x d x + \int 4 d x

\int 3 x^{3} d x = \frac{3 x ^{4}}{4}

\int 5 x^{2} d x = \frac{5 x ^{3}}{3}

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

\int 4 d x = 4 x

= \frac{3 x ^{4}}{4} - \frac{5 x ^{3}}{3} + \frac{3 x ^{2}}{2} + 4 x

Füge eine Konstante zur Lösung hinzu

= \frac{3 x ^{4}}{4} - \frac{5 x ^{3}}{3} + \frac{3 x ^{2}}{2} + 4 x + C

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

f (θ) = e^{θ}

\frac{d}{d x} (cos (e^{4 x}))

d er i v a t i v e f (x) = \frac{7 x ^{2} - 6 x + 1}{3 x + 2}

\frac{d ^{2}}{d x ^{2}} (3 + 5 x + 2 x^{2} + 0.4 x^{3})