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受欢迎的微积分 >

integral of sqrt((x-2)/(x+2))

解答

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

解答

8 - \frac{1}{4} ln \frac{x - 2}{x + 2} + 1 - \frac{1}{4 ( \frac{x - 2}{x + 2} + 1 )} + \frac{1}{4} ln \frac{x - 2}{x + 2} - 1 - \frac{1}{4 ( \frac{x - 2}{x + 2} - 1 )} + C

求解步骤

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

使用换元积分法

= \int \frac{8 u ^{2}}{- u ^{2} + 1 ( - u ^{2} + 1 ) ^{\frac{3}{2}}} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int \frac{u ^{2}}{- u ^{2} + 1 ( - u ^{2} + 1 ) ^{\frac{3}{2}}} d u

(- u^{2} + 1)^{\frac{3}{2}} = (- u^{2} + 1) - u^{2} + 1

= 8 \cdot \int \frac{u ^{2}}{- u ^{2} + 1 ( - u ^{2} + 1 ) - u ^{2} + 1} d u

- u^{2} + 1 (- u^{2} + 1) - u^{2} + 1 = (- u^{2} + 1)^{2}

= 8 \cdot \int \frac{u ^{2}}{( - u ^{2} + 1 ) ^{2}} d u

将

\frac{u ^{2}}{( - u ^{2} + 1 ) ^{2}}

用部份分式展开:

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

= 8 \cdot \int - \frac{1}{4 ( u + 1 )} + \frac{1}{4 ( u + 1 ) ^{2}} + \frac{1}{4 ( u - 1 )} + \frac{1}{4 ( u - 1 ) ^{2}} d u

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

= 8 (- \int \frac{1}{4 ( u + 1 )} d u + \int \frac{1}{4 ( u + 1 ) ^{2}} d u + \int \frac{1}{4 ( u - 1 )} d u + \int \frac{1}{4 ( u - 1 ) ^{2}} d u)

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

\int \frac{1}{4 ( u + 1 ) ^{2}} d u = - \frac{1}{4 ( u + 1 )}

\int \frac{1}{4 ( u - 1 )} d u = \frac{1}{4} ln ∣ u - 1 ∣

\int \frac{1}{4 ( u - 1 ) ^{2}} d u = - \frac{1}{4 ( u - 1 )}

= 8 (- \frac{1}{4} ln ∣ u + 1 ∣ - \frac{1}{4 ( u + 1 )} + \frac{1}{4} ln ∣ u - 1 ∣ - \frac{1}{4 ( u - 1 )})

u = \frac{x - 2}{x + 2}

代回

= 8 - \frac{1}{4} ln \frac{x - 2}{x + 2} + 1 - \frac{1}{4 ( \frac{x - 2}{x + 2} + 1 )} + \frac{1}{4} ln \frac{x - 2}{x + 2} - 1 - \frac{1}{4 ( \frac{x - 2}{x + 2} - 1 )}

解答补常数

= 8 - \frac{1}{4} ln \frac{x - 2}{x + 2} + 1 - \frac{1}{4 ( \frac{x - 2}{x + 2} + 1 )} + \frac{1}{4} ln \frac{x - 2}{x + 2} - 1 - \frac{1}{4 ( \frac{x - 2}{x + 2} - 1 )} + C

作图

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