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受欢迎的微积分 >

integral of (sqrt(9-x^2))/(3x)

解答

\int \frac{9 - x ^{2}}{3 x} d x

解答

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

求解步骤

\int \frac{9 - x ^{2}}{3 x} d x

提出常数:

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

= \frac{1}{3} \cdot \int \frac{9 - x ^{2}}{x} d x

使用换元积分法

= \frac{1}{3} \cdot \int - \frac{u ^{2}}{- 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}{3} (- \int \frac{u ^{2}}{- u ^{2} + 9} d u)

使用换元积分法

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

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

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

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

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

\int 1 d v = v

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

代回

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

化简

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

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

解答补常数

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

作图

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