integral of (5x+3)/(x^2-1)

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

\int \frac{5 x}{x ^{2} - 1} d x = \frac{5}{2} ln x^{2} - 1

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{3 x + 2}{x ^{2} + 9} d x

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

\int 8 x sin (5 - 4 x) d x

\int 3 (x + 4)^{2} d x

in t e g r a l x^{2} ln (x)