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

Lösung

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

ln 4 x^{2} + 4 x + 20 - \frac{8}{19} arctan (\frac{1}{19} (2 x + 1)) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

\int \frac{2 x}{x ^{2} + x + 5} d x = ln 4 x^{2} + 4 x + 20 - \frac{2}{19} arctan (\frac{2 x + 1}{19})

\int \frac{3}{x ^{2} + x + 5} d x = \frac{6}{19} arctan (\frac{2 x + 1}{19})

= ln 4 x^{2} + 4 x + 20 - \frac{2}{19} arctan (\frac{2 x + 1}{19}) - \frac{6}{19} arctan (\frac{2 x + 1}{19})

Vereinfache

ln 4 x^{2} + 4 x + 20 - \frac{2}{19} arctan (\frac{2 x + 1}{19}) - \frac{6}{19} arctan (\frac{2 x + 1}{19}) : ln 4 x^{2} + 4 x + 20 - \frac{8}{19} arctan (\frac{1}{19} (2 x + 1))

= ln 4 x^{2} + 4 x + 20 - \frac{8}{19} arctan (\frac{1}{19} (2 x + 1))

Füge eine Konstante zur Lösung hinzu

= ln 4 x^{2} + 4 x + 20 - \frac{8}{19} arctan (\frac{1}{19} (2 x + 1)) + C