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

解答

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

\frac{x ^{2}}{2} - 2 x + ln ∣ x ∣ + C

求解步骤

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

乘开

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

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

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

\int 2 d x = 2 x

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{x ^{2}}{2} - 2 x + ln ∣ x ∣

解答补常数

= \frac{x ^{2}}{2} - 2 x + ln ∣ x ∣ + C

\frac{\partial}{\partial x} (e^{- 1})

x \to 3 lim (x^{2} - 6 x + 8)

\int \frac{1}{t ^{2} 4 - t ^{2}} d t

\frac{\partial}{\partial x} (- 13 e^{x} sin (y) - 14 y)

\frac{d}{d a} (sin (a) - sin (6 a) + sin (5 a))