integral of (x^2+x-6)/(x^2-4x+3)

Lösung

\int \frac{x ^{2} + x - 6}{x ^{2} - 4 x + 3} d x

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

Schritte zur Lösung

\int \frac{x ^{2} + x - 6}{x ^{2} - 4 x + 3} d x

Multipliziere aus

\frac{x ^{2} + x - 6}{x ^{2} - 4 x + 3} : \frac{x ^{2}}{x ^{2} - 4 x + 3} + \frac{x}{x ^{2} - 4 x + 3} - \frac{6}{x ^{2} - 4 x + 3}

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 ^{2} - 4 x + 3} d x + \int \frac{x}{x ^{2} - 4 x + 3} d x - \int \frac{6}{x ^{2} - 4 x + 3} d x

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

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

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{\partial}{\partial x} (\frac{4 x + y}{x - y})

x \to - \infty lim (3^{4 x})

\int_{0}^{0.5} 5 arccos (x) d x

a re a y = \frac{1}{x ^{2}} + x, x = 2, x = 4

x \to 0 lim (x^{6} sin (x))