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

Lösung

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

\frac{7}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3 + 3 arctan (\frac{x}{3}) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

\int \frac{3 x}{x ^{3} + 3 x} d x = 3 arctan (\frac{x}{3})

\int \frac{7}{x ^{3} + 3 x} d x = 7 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3)

= ln x^{2} + 3 + 3 arctan (\frac{x}{3}) + 7 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3)

Vereinfache

ln x^{2} + 3 + 3 arctan (\frac{x}{3}) + 7 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3) : \frac{7}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3 + 3 arctan (\frac{x}{3})

= \frac{7}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3 + 3 arctan (\frac{x}{3})

Füge eine Konstante zur Lösung hinzu

= \frac{7}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3 + 3 arctan (\frac{x}{3}) + C

\frac{d y}{d x} = (cos (x)) e^{y + s i n (x)}, y (0) = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = - \frac{1}{5}, y^{^{'}} (0) = - \frac{8}{5}

l a pl a ce t r an s f or m 3 e^{2 t} sin (4 t)

\frac{\partial}{\partial β} (sin (α) cos (β))

\frac{\partial}{\partial y} (x^{y^{z}})