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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

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

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

\int y 4 - y^{2} d y

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