integral of (9x^2-8x+24)/(x^3+4x)

Lösung

\int \frac{9 x ^{2} - 8 x + 24}{x ^{3} + 4 x} d x

\frac{3}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 6 ln ∣ x ∣ + C

Schritte zur Lösung

\int \frac{9 x ^{2} - 8 x + 24}{x ^{3} + 4 x} d x

Multipliziere aus

\frac{9 x ^{2} - 8 x + 24}{x ^{3} + 4 x} : \frac{9 x ^{2}}{x ^{3} + 4 x} - \frac{8 x}{x ^{3} + 4 x} + \frac{24}{x ^{3} + 4 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{9 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{8 x}{x ^{3} + 4 x} d x + \int \frac{24}{x ^{3} + 4 x} d x

\int \frac{9 x ^{2}}{x ^{3} + 4 x} d x = \frac{9}{2} ln x^{2} + 4

\int \frac{8 x}{x ^{3} + 4 x} d x = 4 arctan (\frac{x}{2})

\int \frac{24}{x ^{3} + 4 x} d x = 24 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

= \frac{9}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 24 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Vereinfache

\frac{9}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 24 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) : \frac{3}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 6 ln ∣ x ∣

= \frac{3}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 6 ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= \frac{3}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 6 ln ∣ x ∣ + C

x y^{^{'}} - y = x^{2} + 2 x + 1

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

in v erse l a pl a ce \frac{6 ( s + 34 )}{s ( s ^{2} + 10 s + 34 )}

n or ma l f (x) = x^{4} + 5 e^{x}, (0, 5)

d er i v a t i v e f (x) = a x