integral of (4x-5)(2x+1)

解答

\int (4 x - 5) (2 x + 1) d x

4 x^{3} - \frac{4 x ^{3}}{3} - 3 x^{2} - 5 x + C

求解步骤

\int (4 x - 5) (2 x + 1) d x

使用分布积分法

= (2 x + 1) (2 x^{2} - 5 x) - \int 2 (2 x^{2} - 5 x) d x

\int 2 (2 x^{2} - 5 x) d x = \frac{4 x ^{3}}{3} - 5 x^{2}

= (2 x + 1) (2 x^{2} - 5 x) - (\frac{4 x ^{3}}{3} - 5 x^{2})

化简

(2 x + 1) (2 x^{2} - 5 x) - (\frac{4 x ^{3}}{3} - 5 x^{2}) : 4 x^{3} - \frac{4 x ^{3}}{3} - 3 x^{2} - 5 x

= 4 x^{3} - \frac{4 x ^{3}}{3} - 3 x^{2} - 5 x

解答补常数

= 4 x^{3} - \frac{4 x ^{3}}{3} - 3 x^{2} - 5 x + C

\int x^{2} (x^{3} - 4)^{5} d x

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