integral of (x-1)/(4x^2+4x+5)

Lösung

\int \frac{x - 1}{4 x ^{2} + 4 x + 5} d x

\frac{1}{8} ln 4 x^{2} + 4 x + 5 - \frac{3}{8} arctan (x + \frac{1}{2}) + C

Schritte zur Lösung

\int \frac{x - 1}{4 x ^{2} + 4 x + 5} d x

Multipliziere aus

\frac{x - 1}{4 x ^{2} + 4 x + 5} : \frac{x}{4 x ^{2} + 4 x + 5} - \frac{1}{4 x ^{2} + 4 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{x}{4 x ^{2} + 4 x + 5} d x - \int \frac{1}{4 x ^{2} + 4 x + 5} d x

\int \frac{x}{4 x ^{2} + 4 x + 5} d x = \frac{1}{8} (ln 4 x^{2} + 4 x + 5 - arctan (x + \frac{1}{2}))

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

= \frac{1}{8} (ln 4 x^{2} + 4 x + 5 - arctan (x + \frac{1}{2})) - \frac{1}{4} arctan (x + \frac{1}{2})

Vereinfache

\frac{1}{8} (ln 4 x^{2} + 4 x + 5 - arctan (x + \frac{1}{2})) - \frac{1}{4} arctan (x + \frac{1}{2}) : \frac{1}{8} ln 4 x^{2} + 4 x + 5 - \frac{3}{8} arctan (x + \frac{1}{2})

= \frac{1}{8} ln 4 x^{2} + 4 x + 5 - \frac{3}{8} arctan (x + \frac{1}{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{8} ln 4 x^{2} + 4 x + 5 - \frac{3}{8} arctan (x + \frac{1}{2}) + C

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