integral of (x+3)/(x^2-9)

Lösung

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

\frac{1}{2} ln x^{2} - 9 - \frac{1}{2} ln \frac{x}{3} + 1 + \frac{1}{2} ln \frac{x}{3} - 1 + C

Schritte zur Lösung

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

Multipliziere aus

\frac{x + 3}{x ^{2} - 9} : \frac{x}{x ^{2} - 9} + \frac{3}{x ^{2} - 9}

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

\int \frac{x}{x ^{2} - 9} d x = \frac{1}{2} ln x^{2} - 9

\int \frac{3}{x ^{2} - 9} d x = - \frac{1}{2} ln \frac{x}{3} + 1 + \frac{1}{2} ln \frac{x}{3} - 1

= \frac{1}{2} ln x^{2} - 9 - \frac{1}{2} ln \frac{x}{3} + 1 + \frac{1}{2} ln \frac{x}{3} - 1

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln x^{2} - 9 - \frac{1}{2} ln \frac{x}{3} + 1 + \frac{1}{2} ln \frac{x}{3} - 1 + C

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