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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

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

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

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