integral of (7x^2+3x+8)/(x^3-1)

Lösung

\int \frac{7 x ^{2} + 3 x + 8}{x ^{3} - 1} d x

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

Schritte zur Lösung

\int \frac{7 x ^{2} + 3 x + 8}{x ^{3} - 1} d x

Ermittle den Partialbruch von

\frac{7 x ^{2} + 3 x + 8}{x ^{3} - 1} : \frac{6}{x - 1} + \frac{x - 2}{x ^{2} + x + 1}

= \int \frac{6}{x - 1} + \frac{x - 2}{x ^{2} + 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{6}{x - 1} d x + \int \frac{x - 2}{x ^{2} + x + 1} d x

\int \frac{6}{x - 1} d x = 6 ln ∣ x - 1 ∣

\int \frac{x - 2}{x ^{2} + x + 1} d x = \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{5}{3} arctan (\frac{1}{3} (2 x + 1))

= 6 ln ∣ x - 1 ∣ + \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{5}{3} arctan (\frac{1}{3} (2 x + 1))

Füge eine Konstante zur Lösung hinzu

= 6 ln ∣ x - 1 ∣ + \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{5}{3} arctan (\frac{1}{3} (2 x + 1)) + C

\int x \cdot ln (x^{2} + 1) d x

\int \frac{x ^{2} + 12 x - 12}{x ^{3} - 4 x} d x

\int \frac{x ^{2} - 2 x + 4}{x ^{3} + 4 x} d x

\int tan^{3} (x) sin^{2} (x) d x

\int (t^{3} - \frac{e ^{- t} + 4}{e ^{- t}}) d t