integral of (2x-1/(8x))^2

解答

\int (2 x - \frac{1}{8 x})^{2} d x

\frac{4 x ^{3}}{3} - \frac{1}{2} x - \frac{1}{64 x} + C

求解步骤

\int (2 x - \frac{1}{8 x})^{2} d x

乘开

(2 x - \frac{1}{8 x})^{2} : 4 x^{2} - \frac{1}{2} + \frac{1}{64 x ^{2}}

= \int 4 x^{2} - \frac{1}{2} + \frac{1}{64 x ^{2}} 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 4 x^{2} d x - \int \frac{1}{2} d x + \int \frac{1}{64 x ^{2}} d x

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

\int \frac{1}{2} d x = \frac{1}{2} x

\int \frac{1}{64 x ^{2}} d x = - \frac{1}{64 x}

= \frac{4 x ^{3}}{3} - \frac{1}{2} x - \frac{1}{64 x}

解答补常数

= \frac{4 x ^{3}}{3} - \frac{1}{2} x - \frac{1}{64 x} + C

\frac{d}{d x} (\frac{x ^{3}}{9} + \frac{3}{4 x})

a re a 9 (x^{3} - x), 0

\int y^{2} 5 2 - 4 y^{3} d y

s l o p e (- \frac{3}{4}, \frac{1}{4}), (\frac{1}{2}, - \frac{3}{2})

a re a f (x) = x^{2} - 4, g (x) = 0, x = - 4, x = 3