integral of (6x^2)(e^{x^3+2})

Lösung

\int (6 x^{2}) (e^{x^{3} + 2}) d x

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

Schritte zur Lösung

\int 6 x^{2} e^{x^{3} + 2} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int x^{2} e^{x^{3} + 2} d x

Wende die partielle Integration an

= 6 (\frac{1}{3} e^{x^{3} + 2} x^{3} - \int e^{x^{3} + 2} x^{5} d x)

\int e^{x^{3} + 2} x^{5} d x = \frac{e ^{2}}{3} (e^{x^{3}} x^{3} - e^{x^{3}})

= 6 (\frac{1}{3} e^{x^{3} + 2} x^{3} - \frac{e ^{2}}{3} (e^{x^{3}} x^{3} - e^{x^{3}}))

Vereinfache

6 (\frac{1}{3} e^{x^{3} + 2} x^{3} - \frac{e ^{2}}{3} (e^{x^{3}} x^{3} - e^{x^{3}})) : 2 e^{2 + x^{3}}

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

Füge eine Konstante zur Lösung hinzu

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

d er i v a t i v e 9 cos (x)

a re a y = - x^{2} + 16, y = 0, x = - 2, x = 3

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

\frac{d y}{d x} = 9 x

\int \frac{y}{1 + y} d y