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

Lösung

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

\frac{x ^{3}}{3} - \frac{2 x ^{5}}{5} + \frac{x ^{7}}{7} + C

Schritte zur Lösung

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

Multipliziere aus

(1 - x^{2})^{2} x^{2} : x^{2} - 2 x^{4} + x^{6}

= \int x^{2} - 2 x^{4} + x^{6} 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 2 x^{4} d x + \int x^{6} d x

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

\int 2 x^{4} d x = \frac{2 x ^{5}}{5}

\int x^{6} d x = \frac{x ^{7}}{7}

= \frac{x ^{3}}{3} - \frac{2 x ^{5}}{5} + \frac{x ^{7}}{7}

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{3}}{3} - \frac{2 x ^{5}}{5} + \frac{x ^{7}}{7} + C

a re a y = 3 x^{4} - 3 x^{2}, y = 8 x^{2}

d er i v a t i v e y = (2 x - 3)^{4} (x^{2} + x + 1)^{5}

\frac{d}{d x} (a^{6 x^{2}})

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

x^{2} y^{^{'}} + 3 x y = e^{x}