integral of ((x^2-x+12))/(x^3+6x)

Lösung

\int \frac{( x ^{2} - x + 12 )}{x ^{3} + 6 x} d x

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

Schritte zur Lösung

\int \frac{x ^{2} - x + 12}{x ^{3} + 6 x} d x

Ermittle den Partialbruch von

\frac{x ^{2} - x + 12}{x ^{3} + 6 x} : \frac{2}{x} + \frac{- x - 1}{x ^{2} + 6}

= \int \frac{2}{x} + \frac{- x - 1}{x ^{2} + 6} 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{2}{x} d x + \int \frac{- x - 1}{x ^{2} + 6} d x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

Füge eine Konstante zur Lösung hinzu

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

y^{^{'}} + 2 x y = e^{x}

x \to - 6 lim (x + 1)

\int \frac{1}{y ^{2} 5 - y ^{2}} d y

\int_{1}^{e} \frac{4}{z} d z

\frac{d ^{2}}{d x ^{2}} y + 64 y = 0