integral of 2pix(x^2-1)^2

Lösung

\int 2 π x (x^{2} - 1)^{2} d x

\frac{π x ^{6}}{3} - π x^{4} + π x^{2} + C

Schritte zur Lösung

\int 2 π x (x^{2} - 1)^{2} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 π \cdot \int x (x^{2} - 1)^{2} d x

Multipliziere aus

x (x^{2} - 1)^{2} : x^{5} - 2 x^{3} + x

= 2 π \cdot \int x^{5} - 2 x^{3} + x 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

= 2 π (\int x^{5} d x - \int 2 x^{3} d x + \int x d x)

\int x^{5} d x = \frac{x ^{6}}{6}

\int 2 x^{3} d x = \frac{x ^{4}}{2}

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

= 2 π (\frac{x ^{6}}{6} - \frac{x ^{4}}{2} + \frac{x ^{2}}{2})

Vereinfache

2 π (\frac{x ^{6}}{6} - \frac{x ^{4}}{2} + \frac{x ^{2}}{2}) : \frac{π x ^{6}}{3} - π x^{4} + π x^{2}

= \frac{π x ^{6}}{3} - π x^{4} + π x^{2}

Füge eine Konstante zur Lösung hinzu

= \frac{π x ^{6}}{3} - π x^{4} + π x^{2} + C

\int_{4}^{9} \frac{4}{2 x} d x

d er i v a t i v e f (x) = 7 arctan (7 x), x = 1

n = 0 \sum \infty n^{2} (\frac{1}{3})^{n}

\int e^{- s t} t^{5} d t

f^{^{'}} (x) = \frac{1}{3 x ^{\frac{2}{3}}}