integral of (x^2-2x)e^{kx}

解答

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

e^{k x} \frac{1}{k} x^{2} - \frac{2}{k} (e^{k x} \frac{1}{k} x - e^{k x} \frac{1}{k ^{2}}) - 2 (e^{k x} \frac{1}{k} x - e^{k x} \frac{1}{k ^{2}}) + C

求解步骤

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

乘开

(x^{2} - 2 x) e^{k x} : e^{k x} x^{2} - 2 e^{k x} x

= \int e^{k x} x^{2} - 2 e^{k x} x d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int e^{k x} x^{2} d x - \int 2 e^{k x} x d x

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

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

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

解答补常数

= e^{k x} \frac{1}{k} x^{2} - \frac{2}{k} (e^{k x} \frac{1}{k} x - e^{k x} \frac{1}{k ^{2}}) - 2 (e^{k x} \frac{1}{k} x - e^{k x} \frac{1}{k ^{2}}) + C

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