integral of (x-2)(4x+2)

解答

\int (x - 2) (4 x + 2) d x

\frac{4 x ^{3}}{3} - 3 x^{2} - 4 x + C

求解步骤

\int (x - 2) (4 x + 2) d x

(x - 2) (4 x + 2) = 2 (x - 2) (2 x + 1)

= \int 2 (x - 2) (2 x + 1) d x

提出常数:

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

= 2 \cdot \int (x - 2) (2 x + 1) d x

乘开

(x - 2) (2 x + 1) : 2 x^{2} - 3 x - 2

= 2 \cdot \int 2 x^{2} - 3 x - 2 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

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

\int 2 x^{2} d x = \frac{2 x ^{3}}{3}

\int 3 x d x = \frac{3 x ^{2}}{2}

\int 2 d x = 2 x

= 2 (\frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} - 2 x)

化简

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

= \frac{4 x ^{3}}{3} - 3 x^{2} - 4 x

解答补常数

= \frac{4 x ^{3}}{3} - 3 x^{2} - 4 x + C

\int 4.4 e^{(- 3 x + 4)} d x

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