integral of (-x+3)^2

解答

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

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

求解步骤

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

乘开

(- x + 3)^{2} : x^{2} - 6 x + 9

= \int x^{2} - 6 x + 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 x^{2} d x - \int 6 x d x + \int 9 d x

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

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

\int 9 d x = 9 x

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

解答补常数

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

(ln (ln (x)))^{^{'}}

d er i v a t i v e f (x) = x^{3} (3 - x)^{2}

d er i v a t i v e y = cot^{2} (sin (x))

\frac{\partial}{\partial x} ((\frac{e}{y})^{x})

\int_{0}^{4} (3 t^{4} + 3 t + 2) d t