integral of (x^2)/((x^2+2x+5)(x^2-1))

Lösung

\int \frac{x ^{2}}{( x ^{2} + 2 x + 5 ) ( x ^{2} - 1 )} d x

\frac{1}{32} (ln x^{2} + 2 x + 5 + 14 arctan (\frac{x + 1}{2})) - \frac{1}{8} ln ∣ x + 1 ∣ + \frac{1}{16} ln ∣ x - 1 ∣ + C

Schritte zur Lösung

\int \frac{x ^{2}}{( x ^{2} + 2 x + 5 ) ( x ^{2} - 1 )} d x

Ermittle den Partialbruch von

\frac{x ^{2}}{( x ^{2} + 2 x + 5 ) ( x ^{2} - 1 )} : \frac{x + 15}{16 ( x ^{2} + 2 x + 5 )} - \frac{1}{8 ( x + 1 )} + \frac{1}{16 ( x - 1 )}

= \int \frac{x + 15}{16 ( x ^{2} + 2 x + 5 )} - \frac{1}{8 ( x + 1 )} + \frac{1}{16 ( x - 1 )} 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{x + 15}{16 ( x ^{2} + 2 x + 5 )} d x - \int \frac{1}{8 ( x + 1 )} d x + \int \frac{1}{16 ( x - 1 )} d x

\int \frac{x + 15}{16 ( x ^{2} + 2 x + 5 )} d x = \frac{1}{32} (ln x^{2} + 2 x + 5 + 14 arctan (\frac{x + 1}{2}))

\int \frac{1}{8 ( x + 1 )} d x = \frac{1}{8} ln ∣ x + 1 ∣

\int \frac{1}{16 ( x - 1 )} d x = \frac{1}{16} ln ∣ x - 1 ∣

= \frac{1}{32} (ln x^{2} + 2 x + 5 + 14 arctan (\frac{x + 1}{2})) - \frac{1}{8} ln ∣ x + 1 ∣ + \frac{1}{16} ln ∣ x - 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{32} (ln x^{2} + 2 x + 5 + 14 arctan (\frac{x + 1}{2})) - \frac{1}{8} ln ∣ x + 1 ∣ + \frac{1}{16} ln ∣ x - 1 ∣ + C

\int sin^{3} (2 x) (1 - 2 cos (2 x))^{2} d x

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