integral of (x^2-9x+26)/(x^3+x^2+4x+12)

Lösung

\int \frac{x ^{2} - 9 x + 26}{x ^{3} + x ^{2} + 4 x + 12} d x

4 ln ∣ x + 2 ∣ - \frac{3}{2} ln 4 x^{2} - 4 x + 24 - \frac{1}{23} arctan (\frac{1}{23} (2 x - 1)) + C

Schritte zur Lösung

\int \frac{x ^{2} - 9 x + 26}{x ^{3} + x ^{2} + 4 x + 12} d x

Ermittle den Partialbruch von

\frac{x ^{2} - 9 x + 26}{x ^{3} + x ^{2} + 4 x + 12} : \frac{4}{x + 2} + \frac{- 3 x + 1}{x ^{2} - x + 6}

= \int \frac{4}{x + 2} + \frac{- 3 x + 1}{x ^{2} - x + 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{4}{x + 2} d x + \int \frac{- 3 x + 1}{x ^{2} - x + 6} d x

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

\int \frac{- 3 x + 1}{x ^{2} - x + 6} d x = - \frac{3}{2} ln 4 x^{2} - 4 x + 24 - \frac{1}{23} arctan (\frac{1}{23} (2 x - 1))

= 4 ln ∣ x + 2 ∣ - \frac{3}{2} ln 4 x^{2} - 4 x + 24 - \frac{1}{23} arctan (\frac{1}{23} (2 x - 1))

Füge eine Konstante zur Lösung hinzu

= 4 ln ∣ x + 2 ∣ - \frac{3}{2} ln 4 x^{2} - 4 x + 24 - \frac{1}{23} arctan (\frac{1}{23} (2 x - 1)) + C

\int \frac{3 e ^{a}}{4 - e ^{2 a}} d a

\int \frac{1}{2 - 4 x} d x

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

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

\int \frac{3 - z}{2 + z} d z