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受欢迎的微积分 >

integral of 1/(2xsqrt(9-x^2))

解答

\int \frac{1}{2 x 9 - x ^{2}} d x

解答

- \frac{1}{12} (ln \frac{9 - x ^{2}}{3} + 1 - ln \frac{9 - x ^{2}}{3} - 1) + C

求解步骤

\int \frac{1}{2 x 9 - x ^{2}} 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 9 - x ^{2}} d x

使用换元积分法

= \frac{1}{2} \cdot \int - \frac{1}{- u ^{2} + 9} 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{1}{- u ^{2} + 9} d u)

使用换元积分法

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

提出常数:

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

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

使用积分常见定则:

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

= \frac{1}{2} (- \frac{1}{3} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

代回

= \frac{1}{2} - \frac{1}{3} \frac{ln \frac{9 - x ^{2}}{3} + 1}{2} - \frac{ln \frac{9 - x ^{2}}{3} - 1}{2}

化简

\frac{1}{2} - \frac{1}{3} \frac{ln \frac{9 - x ^{2}}{3} + 1}{2} - \frac{ln \frac{9 - x ^{2}}{3} - 1}{2} : - \frac{1}{12} (ln \frac{9 - x ^{2}}{3} + 1 - ln \frac{9 - x ^{2}}{3} - 1)

= - \frac{1}{12} (ln \frac{9 - x ^{2}}{3} + 1 - ln \frac{9 - x ^{2}}{3} - 1)

解答补常数

= - \frac{1}{12} (ln \frac{9 - x ^{2}}{3} + 1 - ln \frac{9 - x ^{2}}{3} - 1) + C

作图

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