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

解答

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

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

求解步骤

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

乘开

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

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

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

\int 1 d x = x

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

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

解答补常数

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

\int_{0}^{\frac{π}{2}} - 10 cos^{5} (x) d x

\frac{\partial}{\partial x} (\frac{2 y}{x ^{2} + y ^{2}})

\int_{- 3}^{4} \frac{1}{x} d x

11 y^{^{''}} + 12 y^{^{'}} + 96 y = 5 cos (4 t)

\frac{\partial}{\partial y} (x^{2} sin (4 y))