integral of e^{2x}x^3

Lösung

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

\frac{1}{2} e^{2 x} x^{3} - \frac{3}{2} (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})) + C

Schritte zur Lösung

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

Wende die partielle Integration an

= \frac{1}{2} e^{2 x} x^{3} - \int \frac{3}{2} e^{2 x} x^{2} d x

\int \frac{3}{2} e^{2 x} x^{2} d x = \frac{3}{2} (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}))

= \frac{1}{2} e^{2 x} x^{3} - \frac{3}{2} (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} e^{2 x} x^{3} - \frac{3}{2} (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})) + C

in v erse l a pl a ce \frac{( s - 1 )}{4 s ^{2} + s + 24}

\frac{d}{d x} (\frac{2 x ^{3} - 3 x ^{2}}{( x - 1 ) ^{2}})

y^{^{'}} = \frac{4 t ^{3} + 1}{2 y - 6}, y (1) = 2

\int \frac{e ^{\frac{6}{t}}}{t ^{2}} d t

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{2}} - 3 x z)