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受欢迎的微积分 >

integral of 1/x*sqrt((x-a)/(x+a))

解答

\int \frac{1}{x} \cdot \frac{x - a}{x + a} d x

解答

ln \frac{x - a}{x + a} + 1 - ln \frac{x - a}{x + a} - 1 - 2 arctan (\frac{x - a}{x + a}) + C

求解步骤

\int \frac{1}{x} \frac{x - a}{x + a} d x

使用换元积分法

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

提出常数:

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

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

将

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

用部份分式展开:

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

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

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

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

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

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

= 4 (\frac{1}{4} ln ∣ u + 1 ∣ - \frac{1}{4} ln ∣ u - 1 ∣ - \frac{1}{2} arctan (u))

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

代回

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

化简

4 (\frac{1}{4} ln \frac{x - a}{x + a} + 1 - \frac{1}{4} ln \frac{x - a}{x + a} - 1 - \frac{1}{2} arctan (\frac{x - a}{x + a})) : ln \frac{x - a}{x + a} + 1 - ln \frac{x - a}{x + a} - 1 - 2 arctan (\frac{x - a}{x + a})

= ln \frac{x - a}{x + a} + 1 - ln \frac{x - a}{x + a} - 1 - 2 arctan (\frac{x - a}{x + a})

解答补常数

= ln \frac{x - a}{x + a} + 1 - ln \frac{x - a}{x + a} - 1 - 2 arctan (\frac{x - a}{x + a}) + C

作图

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