integral of (x-1)/(x^2+4x+29)

Lösung

\int \frac{x - 1}{x ^{2} + 4 x + 29} d x

\frac{1}{2} ln \frac{( x + 2 ) ^{2}}{25} + 1 - \frac{2}{5} arctan (\frac{1}{5} (x + 2)) - \frac{1}{5} arctan (\frac{x + 2}{5}) + C

Schritte zur Lösung

\int \frac{x - 1}{x ^{2} + 4 x + 29} d x

Multipliziere aus

\frac{x - 1}{x ^{2} + 4 x + 29} : \frac{x}{x ^{2} + 4 x + 29} - \frac{1}{x ^{2} + 4 x + 29}

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} + 4 x + 29} d x - \int \frac{1}{x ^{2} + 4 x + 29} d x

\int \frac{x}{x ^{2} + 4 x + 29} d x = \frac{1}{2} ln \frac{( x + 2 ) ^{2}}{25} + 1 - \frac{2}{5} arctan (\frac{1}{5} (x + 2))

\int \frac{1}{x ^{2} + 4 x + 29} d x = \frac{1}{5} arctan (\frac{x + 2}{5})

= \frac{1}{2} ln \frac{( x + 2 ) ^{2}}{25} + 1 - \frac{2}{5} arctan (\frac{1}{5} (x + 2)) - \frac{1}{5} arctan (\frac{x + 2}{5})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln \frac{( x + 2 ) ^{2}}{25} + 1 - \frac{2}{5} arctan (\frac{1}{5} (x + 2)) - \frac{1}{5} arctan (\frac{x + 2}{5}) + C

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