integral of (2x^2+6)/(x^2+2x-8)

Lösung

\int \frac{2 x ^{2} + 6}{x ^{2} + 2 x - 8} d x

2 (x + 1 - \frac{8}{3} ln ∣ x + 4 ∣ + \frac{2}{3} ln ∣ x - 2 ∣) - ln \frac{1}{3} (x + 1) + 1 + ln \frac{1}{3} (x + 1) - 1 + C

Schritte zur Lösung

\int \frac{2 x ^{2} + 6}{x ^{2} + 2 x - 8} d x

Multipliziere aus

\frac{2 x ^{2} + 6}{x ^{2} + 2 x - 8} : \frac{2 x ^{2}}{x ^{2} + 2 x - 8} + \frac{6}{x ^{2} + 2 x - 8}

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 ^{2}}{x ^{2} + 2 x - 8} d x + \int \frac{6}{x ^{2} + 2 x - 8} d x

\int \frac{2 x ^{2}}{x ^{2} + 2 x - 8} d x = 2 (x + 1 - \frac{8}{3} ln ∣ x + 4 ∣ + \frac{2}{3} ln ∣ x - 2 ∣)

\int \frac{6}{x ^{2} + 2 x - 8} d x = - ln \frac{1}{3} (x + 1) + 1 + ln \frac{1}{3} (x + 1) - 1

= 2 (x + 1 - \frac{8}{3} ln ∣ x + 4 ∣ + \frac{2}{3} ln ∣ x - 2 ∣) - ln \frac{1}{3} (x + 1) + 1 + ln \frac{1}{3} (x + 1) - 1

Füge eine Konstante zur Lösung hinzu

= 2 (x + 1 - \frac{8}{3} ln ∣ x + 4 ∣ + \frac{2}{3} ln ∣ x - 2 ∣) - ln \frac{1}{3} (x + 1) + 1 + ln \frac{1}{3} (x + 1) - 1 + C

\int csch (6 x) d x

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