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

Lösung

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

\frac{x ^{3}}{3} - 2 x + 2 arctan (x) + C

Schritte zur Lösung

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

Multipliziere aus

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

\int \frac{x ^{4}}{x ^{2} + 1} d x = \frac{x ^{3}}{3} - x + arctan (x)

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

= \frac{x ^{3}}{3} - x + arctan (x) - (- arctan (x) + x)

Vereinfache

\frac{x ^{3}}{3} - x + arctan (x) - (- arctan (x) + x) : \frac{x ^{3}}{3} - 2 x + 2 arctan (x)

= \frac{x ^{3}}{3} - 2 x + 2 arctan (x)

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{3}}{3} - 2 x + 2 arctan (x) + C

\int cos (x^{2} + 2) d x

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