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Beliebt Rechnen >

integral of (x^2-4x+1)/((x^2-x+1)(x+4))

Lösung

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

Lösung

- \frac{2}{7} ln 4 x^{2} - 4 x + 4 - \frac{2 3}{7} arctan (\frac{1}{3} (2 x - 1)) + \frac{11}{7} ln ∣ x + 4 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{x ^{2} - 4 x + 1}{( x ^{2} - x + 1 ) ( x + 4 )} : \frac{- 4 x - 1}{7 ( x ^{2} - x + 1 )} + \frac{11}{7 ( x + 4 )}

= \int \frac{- 4 x - 1}{7 ( x ^{2} - x + 1 )} + \frac{11}{7 ( x + 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{- 4 x - 1}{7 ( x ^{2} - x + 1 )} d x + \int \frac{11}{7 ( x + 4 )} d x

\int \frac{- 4 x - 1}{7 ( x ^{2} - x + 1 )} d x = \frac{1}{7 3} (- 23 ln 4 x^{2} - 4 x + 4 - 6 arctan (\frac{2 x - 1}{3}))

\int \frac{11}{7 ( x + 4 )} d x = \frac{11}{7} ln ∣ x + 4 ∣

= \frac{1}{7 3} (- 23 ln 4 x^{2} - 4 x + 4 - 6 arctan (\frac{2 x - 1}{3})) + \frac{11}{7} ln ∣ x + 4 ∣

Vereinfache

\frac{1}{7 3} (- 23 ln 4 x^{2} - 4 x + 4 - 6 arctan (\frac{2 x - 1}{3})) + \frac{11}{7} ln ∣ x + 4 ∣ : - \frac{2}{7} ln 4 x^{2} - 4 x + 4 - \frac{2 3}{7} arctan (\frac{1}{3} (2 x - 1)) + \frac{11}{7} ln ∣ x + 4 ∣

= - \frac{2}{7} ln 4 x^{2} - 4 x + 4 - \frac{2 3}{7} arctan (\frac{1}{3} (2 x - 1)) + \frac{11}{7} ln ∣ x + 4 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{2}{7} ln 4 x^{2} - 4 x + 4 - \frac{2 3}{7} arctan (\frac{1}{3} (2 x - 1)) + \frac{11}{7} ln ∣ x + 4 ∣ + C

Graph

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