integral of ((x-3))/(x^2+6x-10)

Lösung

\int \frac{( x - 3 )}{x ^{2} + 6 x - 10} d x

\frac{6 + 19}{2 19} ln x + 3 + 19 + \frac{19 - 6}{2 19} ln x + 3 - 19 + C

Schritte zur Lösung

\int \frac{x - 3}{x ^{2} + 6 x - 10} d x

Vervollständige das Quadrat

x^{2} + 6 x - 10 : (x + 3)^{2} - 19

= \int \frac{x - 3}{( x + 3 ) ^{2} - 19} d x

Wende U-Substitution an

= \int \frac{u - 6}{u ^{2} - 19} d u

Ermittle den Partialbruch von

\frac{u - 6}{u ^{2} - 19} : \frac{6 + 19}{2 19 ( u + 19 )} + \frac{19 - 6}{2 19 ( u - 19 )}

= \int \frac{6 + 19}{2 19 ( u + 19 )} + \frac{19 - 6}{2 19 ( u - 19 )} d u

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{6 + 19}{2 19 ( u + 19 )} d u + \int \frac{19 - 6}{2 19 ( u - 19 )} d u

\int \frac{6 + 19}{2 19 ( u + 19 )} d u = \frac{6 + 19}{2 19} ln u + 19

\int \frac{19 - 6}{2 19 ( u - 19 )} d u = \frac{19 - 6}{2 19} ln u - 19

= \frac{6 + 19}{2 19} ln u + 19 + \frac{19 - 6}{2 19} ln u - 19

Setze in

u = x + 3

ein

= \frac{6 + 19}{2 19} ln x + 3 + 19 + \frac{19 - 6}{2 19} ln x + 3 - 19

Füge eine Konstante zur Lösung hinzu

= \frac{6 + 19}{2 19} ln x + 3 + 19 + \frac{19 - 6}{2 19} ln x + 3 - 19 + C

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