integral of (6x^2-5x+8)/(x^3+4x)

Lösung

\int \frac{6 x ^{2} - 5 x + 8}{x ^{3} + 4 x} d x

- \frac{5}{2} arctan (\frac{x}{2}) + 2 ln x^{2} + 4 + 2 ln ∣ x ∣ + C

Schritte zur Lösung

\int \frac{6 x ^{2} - 5 x + 8}{x ^{3} + 4 x} d x

Multipliziere aus

\frac{6 x ^{2} - 5 x + 8}{x ^{3} + 4 x} : \frac{6 x ^{2}}{x ^{3} + 4 x} - \frac{5 x}{x ^{3} + 4 x} + \frac{8}{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{6 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{5 x}{x ^{3} + 4 x} d x + \int \frac{8}{x ^{3} + 4 x} d x

\int \frac{6 x ^{2}}{x ^{3} + 4 x} d x = 3 ln x^{2} + 4

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

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

= 3 ln x^{2} + 4 - \frac{5}{2} arctan (\frac{x}{2}) + 2 ln ∣ x ∣ - ln x^{2} + 4

Vereinfache

= - \frac{5}{2} arctan (\frac{x}{2}) + 2 ln x^{2} + 4 + 2 ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{5}{2} arctan (\frac{x}{2}) + 2 ln x^{2} + 4 + 2 ln ∣ x ∣ + C

\int_{0}^{4} 16 x^{2} - x^{4} d x

\frac{\partial}{\partial x} (x^{2} sin (6 t))

\frac{d}{d x} (- \frac{e ^{- x} ( x ^{3} + 3 ( x ^{2} + 2 x + 2 ) )}{x ^{4}})

t an g e n t f (x) = \frac{x ^{2} + 8}{x + 28}, (5, 1)

s l o p e in t erce pt (4, - 9), (2, - 2)