integral of x(2x+1)^2

Lösung

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

x^{4} + \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2} + C

Schritte zur Lösung

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

Multipliziere aus

x (2 x + 1)^{2} : 4 x^{3} + 4 x^{2} + x

= \int 4 x^{3} + 4 x^{2} + 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

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

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

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

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

= x^{4} + \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2}

Füge eine Konstante zur Lösung hinzu

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

t an g e n t f (x) = x^{2} + 4, (4, 20)

x \to 2 - lim (e^{\frac{1}{x - 2}})

y^{^{'}} = y (\frac{1}{x} - y)

a re a y = x^{2} - 4 x - 5, y = 2 x - 5

\int_{2.25}^{2.75} 2 d x