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

Lösung

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

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

Schritte zur Lösung

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

Lange Division

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

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

\int x^{2} d x = \frac{x ^{3}}{3}

\int 1 d x = x

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

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

Füge eine Konstante zur Lösung hinzu

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

\int_{1}^{3} (x - 1) ln (x) d x

\frac{d}{d x} (\frac{4}{3} p (x) i r (x) (x)^{3})

\frac{\partial}{\partial x} (y ln (x^{4} + y^{3} + 2) - 7)

d er i v a t i v e (\frac{4}{3}) p i x^{3}

\frac{d}{d x} ((x^{3} + 2 x)^{4})