integral of (16x^3-3x^2+2)

Lösung

\int (16 x^{3} - 3 x^{2} + 2) d x

4 x^{4} - x^{3} + 2 x + C

Schritte zur Lösung

\int 16 x^{3} - 3 x^{2} + 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 16 x^{3} d x - \int 3 x^{2} d x + \int 2 d x

\int 16 x^{3} d x = 4 x^{4}

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

\int 2 d x = 2 x

= 4 x^{4} - x^{3} + 2 x

Füge eine Konstante zur Lösung hinzu

= 4 x^{4} - x^{3} + 2 x + C

x \to 2 lim (\frac{3 x ^{2} + 4 - 4 x ^{2} + 6 x}{x ^{2} - 4})

\int \frac{cos ( x )}{3 + sin ( x )} d x

\int \frac{x ^{3} + 4 x - 2}{( x - 2 ) ( x + 1 )} d x

t an g e n t f (x) = \frac{x ^{3}}{4}, at x = - 6

\int tan^{4} (x) sec^{6} (x) d x