integral of (9x^2-4x+2)

Lösung

\int (9 x^{2} - 4 x + 2) d x

3 x^{3} - 2 x^{2} + 2 x + C

Schritte zur Lösung

\int 9 x^{2} - 4 x + 2 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 9 x^{2} d x - \int 4 x d x + \int 2 d x

\int 9 x^{2} d x = 3 x^{3}

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

\int 2 d x = 2 x

= 3 x^{3} - 2 x^{2} + 2 x

Füge eine Konstante zur Lösung hinzu

= 3 x^{3} - 2 x^{2} + 2 x + C

in v erse l a pl a ce \frac{4}{( s + 1 ) ( s ^{2} + 4 s + 8 )}

x \to - 2 lim (\frac{x - 3}{x + 5})

\frac{\partial}{\partial y} (- x sin (y))

y^{^{'}} = (t + y)^{2} - 1, y (1) = 6

\frac{d}{d x} (4 x + 2 x^{2} sin (\frac{1}{x}))