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

Lösung

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

\frac{3 x ^{4}}{4} + \frac{4 x ^{3}}{3} - 6 x^{2} + 9 x + C

Schritte zur Lösung

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

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

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

= \frac{3 x ^{4}}{4} + \frac{4 x ^{3}}{3} - 6 x^{2} + 9 x

Füge eine Konstante zur Lösung hinzu

= \frac{3 x ^{4}}{4} + \frac{4 x ^{3}}{3} - 6 x^{2} + 9 x + C

\frac{d ^{2}}{d x ^{2}} (e^{6 e^{x}})

\int cos^{2} (x s i) n^{2} x d x

in v erse l a pl a ce \frac{1960}{s ^{2} + 196}

d er i v a t i v e y = sin (tan (6 x))

\frac{\partial}{\partial x} (cos^{3} (x) y - 1)