integral of 1-(x-3)^2

解答

\int 1 - (x - 3)^{2} d x

- \frac{x ^{3}}{3} + 3 x^{2} - 8 x + C

求解步骤

\int 1 - (x - 3)^{2} d x

乘开

1 - (x - 3)^{2} : - x^{2} + 6 x - 8

= \int - x^{2} + 6 x - 8 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 x^{2} d x + \int 6 x d x - \int 8 d x

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

\int 6 x d x = 3 x^{2}

\int 8 d x = 8 x

= - \frac{x ^{3}}{3} + 3 x^{2} - 8 x

解答补常数

= - \frac{x ^{3}}{3} + 3 x^{2} - 8 x + C

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

\int 16 csch^{2} (x) coth (x) d x

\int 4 x^{3} (x^{2} + 1)^{65} d x

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

\int \frac{8}{sec ( x ) - 1} d x