integral of (8x^2-5x+1)/(x^2+x)

Lösung

\int \frac{8 x ^{2} - 5 x + 1}{x ^{2} + x} d x

\frac{8 x ^{2}}{x + 1} - 14 ln ∣ x + 1 ∣ - \frac{8}{x + 1} + ln ∣ x ∣ + C

Schritte zur Lösung

\int \frac{8 x ^{2} - 5 x + 1}{x ^{2} + x} d x

Multipliziere aus

\frac{8 x ^{2} - 5 x + 1}{x ^{2} + x} : \frac{8 x ^{2}}{x ^{2} + x} - \frac{5 x}{x ^{2} + x} + \frac{1}{x ^{2} + 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{8 x ^{2}}{x ^{2} + x} d x - \int \frac{5 x}{x ^{2} + x} d x + \int \frac{1}{x ^{2} + x} d x

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

\int \frac{5 x}{x ^{2} + x} d x = 5 ln ∣ x + 1 ∣

\int \frac{1}{x ^{2} + x} d x = ln ∣ x ∣ - ln ∣ x + 1 ∣

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

Vereinfache

8 (\frac{x ^{2}}{x + 1} - ln ∣ x + 1 ∣ - \frac{1}{x + 1}) - 5 ln ∣ x + 1 ∣ + ln ∣ x ∣ - ln ∣ x + 1 ∣ : \frac{8 x ^{2}}{x + 1} - 14 ln ∣ x + 1 ∣ - \frac{8}{x + 1} + ln ∣ x ∣

= \frac{8 x ^{2}}{x + 1} - 14 ln ∣ x + 1 ∣ - \frac{8}{x + 1} + ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= \frac{8 x ^{2}}{x + 1} - 14 ln ∣ x + 1 ∣ - \frac{8}{x + 1} + ln ∣ x ∣ + C

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