integral of (7x^2+3x)/(x^3+4x^2+3x+12)

Lösung

\int \frac{7 x ^{2} + 3 x}{x ^{3} + 4 x ^{2} + 3 x + 12} d x

\frac{100}{19} ln ∣ x + 4 ∣ + \frac{3}{19} (\frac{11 3}{2} ln \frac{x ^{2}}{3} + 1 - 25 arctan (\frac{x}{3})) + C

Schritte zur Lösung

\int \frac{7 x ^{2} + 3 x}{x ^{3} + 4 x ^{2} + 3 x + 12} d x

Ermittle den Partialbruch von

\frac{7 x ^{2} + 3 x}{x ^{3} + 4 x ^{2} + 3 x + 12} : \frac{100}{19 ( x + 4 )} + \frac{33 x - 75}{19 ( x ^{2} + 3 )}

= \int \frac{100}{19 ( x + 4 )} + \frac{33 x - 75}{19 ( x ^{2} + 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 \frac{100}{19 ( x + 4 )} d x + \int \frac{33 x - 75}{19 ( x ^{2} + 3 )} d x

\int \frac{100}{19 ( x + 4 )} d x = \frac{100}{19} ln ∣ x + 4 ∣

\int \frac{33 x - 75}{19 ( x ^{2} + 3 )} d x = \frac{3}{19} (\frac{11 3}{2} ln \frac{x ^{2}}{3} + 1 - 25 arctan (\frac{x}{3}))

= \frac{100}{19} ln ∣ x + 4 ∣ + \frac{3}{19} (\frac{11 3}{2} ln \frac{x ^{2}}{3} + 1 - 25 arctan (\frac{x}{3}))

Füge eine Konstante zur Lösung hinzu

= \frac{100}{19} ln ∣ x + 4 ∣ + \frac{3}{19} (\frac{11 3}{2} ln \frac{x ^{2}}{3} + 1 - 25 arctan (\frac{x}{3})) + C

\frac{d}{d x} (\frac{( x ^{2} + x ) ( x ^{2} - x + 1 )}{x ^{4}})

n \to \infty lim (- 2^{n})

n \to \infty lim (n - 3)

t an g e n t y = 3 + 4 x^{2} - 2 x^{3}, at x = 1

n = 0 \sum \infty \frac{2 n}{n !}