integral of x^2e^{x+4}

解答

\int x^{2} e^{x + 4} d x

e^{4} (x^{2} e^{x} - 2 (e^{x} x - e^{x})) + C

求解步骤

\int x^{2} e^{x + 4} d x

化简

e^{x + 4} : e^{x} e^{4}

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

提出常数:

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

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

使用分布积分法

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

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

= e^{4} (x^{2} e^{x} - 2 (e^{x} x - e^{x}))

解答补常数

= e^{4} (x^{2} e^{x} - 2 (e^{x} x - e^{x})) + C

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

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

l a pl a ce t r an s f or m e^{(2 t)}

\int \frac{1}{x ^{2} + 10 x} d x

\frac{d y}{d x} = - \frac{( e ^{x} )}{8}