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

Lösung

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

2 ln ∣ x ∣ - \frac{3}{2} ln x^{2} + 4 + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{- x ^{2} + 8}{x ^{3} + 4 x} : \frac{2}{x} - \frac{3 x}{x ^{2} + 4}

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

Füge eine Konstante zur Lösung hinzu

= 2 ln ∣ x ∣ - \frac{3}{2} ln x^{2} + 4 + C

\int (2 x + 4 y) d x

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\int \frac{x}{2 + 4 x ^{2}} d x

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