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

Lösung

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

\frac{7}{2} ln x^{2} - 1 + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + C

Schritte zur Lösung

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

Multipliziere aus

\frac{7 x - 1}{x ^{2} - 1} : \frac{7 x}{x ^{2} - 1} - \frac{1}{x ^{2} - 1}

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

\int \frac{7 x}{x ^{2} - 1} d x = \frac{7}{2} ln x^{2} - 1

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

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

Vereinfache

= \frac{7}{2} ln x^{2} - 1 + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{7}{2} ln x^{2} - 1 + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + C