integral of (-6x^3+9x^2+4x-3)

Lösung

\int (- 6 x^{3} + 9 x^{2} + 4 x - 3) d x

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

Schritte zur Lösung

\int - 6 x^{3} + 9 x^{2} + 4 x - 3 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 6 x^{3} d x + \int 9 x^{2} d x + \int 4 x d x - \int 3 d x

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

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

\int 3 d x = 3 x

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

Füge eine Konstante zur Lösung hinzu

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

\int 2 x e x^{2} d x

\int x 3 - 2 x d x

\int \frac{e ^{2 x} - 1}{e ^{x}} d x

\int \frac{1}{1 + cos ( x ) + sin ( x )} d x

\int 3^{4 - 11 x} d x