integral of (x^3-1)/(x^2+1)

Lösung

\int \frac{x ^{3} - 1}{x ^{2} + 1} d x

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

Schritte zur Lösung

\int \frac{x ^{3} - 1}{x ^{2} + 1} d x

Multipliziere aus

\frac{x ^{3} - 1}{x ^{2} + 1} : \frac{x ^{3}}{x ^{2} + 1} - \frac{1}{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 ^{3}}{x ^{2} + 1} d x - \int \frac{1}{x ^{2} + 1} d x

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

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d x} (cot^{4} (x))

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

\frac{d}{d x} (e^{(x y)})

\int x + 5 d x

d er i v a t i v e \frac{- x ^{2}}{2}