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

Lösung

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

\frac{4 x ^{5}}{5} + \frac{4 x ^{3}}{3} + x + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

\int 1 d x = x

= \frac{4 x ^{5}}{5} + \frac{4 x ^{3}}{3} + x

Füge eine Konstante zur Lösung hinzu

= \frac{4 x ^{5}}{5} + \frac{4 x ^{3}}{3} + x + C

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

t an g e n t f (x) = (\frac{x + 1}{x - 3})^{2}, at x = 7

\frac{\partial}{\partial x} (x^{2} y^{2} cos (x))

\frac{\partial}{\partial x} (ln (1 + x y))

d er i v a t i v e f (x) = \frac{2 x + 3}{3 x + 2}