integral of (2x+1)/(x^2-7x+12)

Lösung

\int \frac{2 x + 1}{x ^{2} - 7 x + 12} d x

- 7 ln ∣ x - 3 ∣ + 9 ln ∣ x - 4 ∣ + C

Schritte zur Lösung

\int \frac{2 x + 1}{x ^{2} - 7 x + 12} d x

Ermittle den Partialbruch von

\frac{2 x + 1}{x ^{2} - 7 x + 12} : - \frac{7}{x - 3} + \frac{9}{x - 4}

= \int - \frac{7}{x - 3} + \frac{9}{x - 4} 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{7}{x - 3} d x + \int \frac{9}{x - 4} d x

\int \frac{7}{x - 3} d x = 7 ln ∣ x - 3 ∣

\int \frac{9}{x - 4} d x = 9 ln ∣ x - 4 ∣

= - 7 ln ∣ x - 3 ∣ + 9 ln ∣ x - 4 ∣

Füge eine Konstante zur Lösung hinzu

= - 7 ln ∣ x - 3 ∣ + 9 ln ∣ x - 4 ∣ + C

d er i v a t i v e y = e^{4 x}

l o n g d i v i s i o n \frac{x ^{2} - x - 2}{x ^{2} - 6 x + 9}

\int_{1}^{2} f (x) d x

\int \frac{1}{64 x ^{2} - 81} d x

n = 0 \sum \infty cos (π \cdot n)