integral of 6x^8-20x^4+x^2+9

解答

\int 6 x^{8} - 20 x^{4} + x^{2} + 9 d x

\frac{2 x ^{9}}{3} - 4 x^{5} + \frac{x ^{3}}{3} + 9 x + C

求解步骤

\int 6 x^{8} - 20 x^{4} + x^{2} + 9 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 6 x^{8} d x - \int 20 x^{4} d x + \int x^{2} d x + \int 9 d x

\int 6 x^{8} d x = \frac{2 x ^{9}}{3}

\int 20 x^{4} d x = 4 x^{5}

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

\int 9 d x = 9 x

= \frac{2 x ^{9}}{3} - 4 x^{5} + \frac{x ^{3}}{3} + 9 x

解答补常数

= \frac{2 x ^{9}}{3} - 4 x^{5} + \frac{x ^{3}}{3} + 9 x + C

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