integral of ((2x+1))/(x^2-9)

Lösung

\int \frac{( 2 x + 1 )}{x ^{2} - 9} d x

ln x^{2} - 9 - \frac{1}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1) + C

Schritte zur Lösung

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

Multipliziere aus

\frac{2 x + 1}{x ^{2} - 9} : \frac{2 x}{x ^{2} - 9} + \frac{1}{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{2 x}{x ^{2} - 9} d x + \int \frac{1}{x ^{2} - 9} d x

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

\int \frac{1}{x ^{2} - 9} d x = - \frac{1}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1)

= ln x^{2} - 9 - \frac{1}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1)

Füge eine Konstante zur Lösung hinzu

= ln x^{2} - 9 - \frac{1}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1) + C

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