integral of (9x^5-6x^3+4x)

Lösung

\int (9 x^{5} - 6 x^{3} + 4 x) d x

\frac{3 x ^{6}}{2} - \frac{3 x ^{4}}{2} + 2 x^{2} + C

Schritte zur Lösung

\int 9 x^{5} - 6 x^{3} + 4 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 9 x^{5} d x - \int 6 x^{3} d x + \int 4 x d x

\int 9 x^{5} d x = \frac{3 x ^{6}}{2}

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

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

= \frac{3 x ^{6}}{2} - \frac{3 x ^{4}}{2} + 2 x^{2}

Füge eine Konstante zur Lösung hinzu

= \frac{3 x ^{6}}{2} - \frac{3 x ^{4}}{2} + 2 x^{2} + C