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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

\frac{u ^{2}}{- u ^{2} + 1} = \frac{1}{- u ^{2} + 1} - 1

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

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

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

\int 1 d u = u

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

u = 1 - x^{2}

代回

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

化简

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

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

解答补常数

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

作图

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