integral of (2x^3+7x^2+2x+9)/(2x+3)

Lösung

\int \frac{2 x ^{3} + 7 x ^{2} + 2 x + 9}{2 x + 3} d x

\frac{x ^{3}}{3} + x^{2} - 2 x + \frac{15}{2} ln ∣ 2 x + 3 ∣ + C

Schritte zur Lösung

\int \frac{2 x ^{3} + 7 x ^{2} + 2 x + 9}{2 x + 3} d x

Lange Division

\frac{2 x ^{3} + 7 x ^{2} + 2 x + 9}{2 x + 3} : x^{2} + 2 x - 2 + \frac{15}{2 x + 3}

= \int x^{2} + 2 x - 2 + \frac{15}{2 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 x^{2} d x + \int 2 x d x - \int 2 d x + \int \frac{15}{2 x + 3} d x

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

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

\int 2 d x = 2 x

\int \frac{15}{2 x + 3} d x = \frac{15}{2} ln ∣ 2 x + 3 ∣

= \frac{x ^{3}}{3} + x^{2} - 2 x + \frac{15}{2} ln ∣ 2 x + 3 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{3}}{3} + x^{2} - 2 x + \frac{15}{2} ln ∣ 2 x + 3 ∣ + C

y^{^{'}} - 2 y = - 4 y^{3}

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

\int (1 + t)^{2} d t

d er i v a t i v e y = \frac{2 x + 4}{1 - x}

\frac{d N}{d t} + N = Nt e^{t + 4}