integral of (3x+1)/(x^2-4x+3)

Lösung

\int \frac{3 x + 1}{x ^{2} - 4 x + 3} d x

- 2 ln ∣ x - 1 ∣ + 5 ln ∣ x - 3 ∣ + C

Schritte zur Lösung

\int \frac{3 x + 1}{x ^{2} - 4 x + 3} d x

Ermittle den Partialbruch von

\frac{3 x + 1}{x ^{2} - 4 x + 3} : - \frac{2}{x - 1} + \frac{5}{x - 3}

= \int - \frac{2}{x - 1} + \frac{5}{x - 3} 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{2}{x - 1} d x + \int \frac{5}{x - 3} d x

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

\int \frac{5}{x - 3} d x = 5 ln ∣ x - 3 ∣

= - 2 ln ∣ x - 1 ∣ + 5 ln ∣ x - 3 ∣

Füge eine Konstante zur Lösung hinzu

= - 2 ln ∣ x - 1 ∣ + 5 ln ∣ x - 3 ∣ + C

\int \frac{5}{x ^{3}} d x

\int 9 x - x^{2} d x

\int (3 x - 2) (x + 3) d x

\int \frac{x ^{4}}{x ^{5} - 8} d x

\int (cos (2 x))^{3} d x