integral of (4x^3+5x)e^{x^2}

Lösung

\int (4 x^{3} + 5 x) e^{x^{2}} d x

2 e^{x^{2}} x^{2} + \frac{1}{2} e^{x^{2}} + C

Schritte zur Lösung

\int (4 x^{3} + 5 x) e^{x^{2}} d x

Multipliziere aus

(4 x^{3} + 5 x) e^{x^{2}} : 4 e^{x^{2}} x^{3} + 5 e^{x^{2}} x

= \int 4 e^{x^{2}} x^{3} + 5 e^{x^{2}} 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 4 e^{x^{2}} x^{3} d x + \int 5 e^{x^{2}} x d x

\int 4 e^{x^{2}} x^{3} d x = 2 (e^{x^{2}} x^{2} - e^{x^{2}})

\int 5 e^{x^{2}} x d x = \frac{5}{2} e^{x^{2}}

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

Vereinfache

2 (e^{x^{2}} x^{2} - e^{x^{2}}) + \frac{5}{2} e^{x^{2}} : 2 e^{x^{2}} x^{2} + \frac{1}{2} e^{x^{2}}

= 2 e^{x^{2}} x^{2} + \frac{1}{2} e^{x^{2}}

Füge eine Konstante zur Lösung hinzu

= 2 e^{x^{2}} x^{2} + \frac{1}{2} e^{x^{2}} + C

\int \frac{x ^{2}}{x ^{3 + 1}} d x

\int \frac{1}{3} e^{t} d t

x^{2} + x \frac{d y}{d x} = y

\frac{\partial}{\partial y} (- 3 y)

y^{^{'}} + \frac{t}{t ^{2} - 16} y = \frac{e ^{t}}{t - 9}