integral of x^2-9x+3

解答

\int x^{2} - 9 x + 3 d x

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

求解步骤

\int x^{2} - 9 x + 3 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 9 x d x + \int 3 d x

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

\int 9 x d x = \frac{9 x ^{2}}{2}

\int 3 d x = 3 x

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

解答补常数

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

n = 1 \sum \infty \frac{9}{6 + 9 ^{n}}

\int \frac{1}{1 - 9 x ^{2}} d x

\int 2 cos (x^{2}) d x

\frac{d}{d x} (\frac{2 x ^{2} + 8 x + 8}{x})

d er i v a t i v e f (x) = (2 e^{x} - 3 x^{5})