integral of 6x^2e^{6x}

解答

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

e^{6 x} x^{2} - \frac{1}{3} e^{6 x} x + \frac{1}{18} e^{6 x} + C

求解步骤

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

提出常数:

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

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

使用分布积分法

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

\int \frac{1}{3} e^{6 x} x d x = \frac{1}{108} (6 e^{6 x} x - e^{6 x})

= 6 (\frac{1}{6} e^{6 x} x^{2} - \frac{1}{108} (6 e^{6 x} x - e^{6 x}))

化简

6 (\frac{1}{6} e^{6 x} x^{2} - \frac{1}{108} (6 e^{6 x} x - e^{6 x})) : e^{6 x} x^{2} - \frac{1}{3} e^{6 x} x + \frac{1}{18} e^{6 x}

= e^{6 x} x^{2} - \frac{1}{3} e^{6 x} x + \frac{1}{18} e^{6 x}

解答补常数

= e^{6 x} x^{2} - \frac{1}{3} e^{6 x} x + \frac{1}{18} e^{6 x} + C

a re a y = x^{3} + x^{2} - 6 x, y = 6 x

l a pl a ce t r an s f or m e^{(3 t)} (4 cos (3 t) - \frac{10}{3} sin (3 t))

x \to - 1 + lim (\frac{x}{∣ x + 1 ∣})

so l v e f or y, y = x e^{y}

\int_{- 3}^{- 2} (2 x (3 - x^{2})^{3}) d x