integral of (x^3-3x+2)/(x(x^2+2x+1))

Lösung

\int \frac{x ^{3} - 3 x + 2}{x ( x ^{2} + 2 x + 1 )} d x

2 x - \frac{x ^{2}}{x + 1} - 4 ln ∣ x + 1 ∣ + \frac{5}{x + 1} + 2 ln ∣ x ∣ + 2 + C

Schritte zur Lösung

\int \frac{x ^{3} - 3 x + 2}{x ( x ^{2} + 2 x + 1 )} d x

Multipliziere aus

\frac{x ^{3} - 3 x + 2}{x ( x ^{2} + 2 x + 1 )} : \frac{x ^{3}}{x ( x ^{2} + 2 x + 1 )} - \frac{3 x}{x ( x ^{2} + 2 x + 1 )} + \frac{2}{x ( x ^{2} + 2 x + 1 )}

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 ^{3}}{x ( x ^{2} + 2 x + 1 )} d x - \int \frac{3 x}{x ( x ^{2} + 2 x + 1 )} d x + \int \frac{2}{x ( x ^{2} + 2 x + 1 )} d x

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

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

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

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

Vereinfache

- \frac{x ^{2}}{x + 1} + 2 (x + 1 - ln ∣ x + 1 ∣) - (- \frac{3}{x + 1}) + 2 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1}) : 2 x - \frac{x ^{2}}{x + 1} - 4 ln ∣ x + 1 ∣ + \frac{5}{x + 1} + 2 ln ∣ x ∣ + 2

= 2 x - \frac{x ^{2}}{x + 1} - 4 ln ∣ x + 1 ∣ + \frac{5}{x + 1} + 2 ln ∣ x ∣ + 2

Füge eine Konstante zur Lösung hinzu

= 2 x - \frac{x ^{2}}{x + 1} - 4 ln ∣ x + 1 ∣ + \frac{5}{x + 1} + 2 ln ∣ x ∣ + 2 + C

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