integral of (x^2-x-6)/(x^3+x)

Lösung

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

\frac{7}{2} ln x^{2} + 1 - arctan (x) - 6 ln ∣ x ∣ + C

Schritte zur Lösung

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

Multipliziere aus

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

\int \frac{x ^{2}}{x ^{3} + x} d x = \frac{1}{2} ln x^{2} + 1

\int \frac{x}{x ^{3} + x} d x = arctan (x)

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

= \frac{1}{2} ln x^{2} + 1 - arctan (x) - 6 (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1)

Vereinfache

\frac{1}{2} ln x^{2} + 1 - arctan (x) - 6 (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1) : \frac{7}{2} ln x^{2} + 1 - arctan (x) - 6 ln ∣ x ∣

= \frac{7}{2} ln x^{2} + 1 - arctan (x) - 6 ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= \frac{7}{2} ln x^{2} + 1 - arctan (x) - 6 ln ∣ x ∣ + C

\int (x^{2} + 3 x^{13})^{5} d x

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