integral of 3x^7+4x^5-2x^3-x+1

解答

\int 3 x^{7} + 4 x^{5} - 2 x^{3} - x + 1 d x

\frac{3 x ^{8}}{8} + \frac{2 x ^{6}}{3} - \frac{x ^{4}}{2} - \frac{x ^{2}}{2} + x + C

求解步骤

\int 3 x^{7} + 4 x^{5} - 2 x^{3} - x + 1 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 3 x^{7} d x + \int 4 x^{5} d x - \int 2 x^{3} d x - \int x d x + \int 1 d x

\int 3 x^{7} d x = \frac{3 x ^{8}}{8}

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

\int 2 x^{3} d x = \frac{x ^{4}}{2}

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

\int 1 d x = x

= \frac{3 x ^{8}}{8} + \frac{2 x ^{6}}{3} - \frac{x ^{4}}{2} - \frac{x ^{2}}{2} + x

解答补常数

= \frac{3 x ^{8}}{8} + \frac{2 x ^{6}}{3} - \frac{x ^{4}}{2} - \frac{x ^{2}}{2} + x + C

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

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

\frac{d}{d x} (\frac{1400}{( 1 + 6 e ^{- 0.02 x} )})

t a y l or f (x) = x^{- 2}

d er i v a t i v e f (x) = \frac{x}{x ^{2} + 5}