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受欢迎的微积分 >

integral of (sqrt(400-x^2))/x

解答

\int \frac{400 - x ^{2}}{x} d x

解答

- 10 ln \frac{1}{20} - x^{2} + 400 + 1 + 10 ln \frac{1}{20} - x^{2} + 400 - 1 + 400 - x^{2} + C

求解步骤

\int \frac{400 - x ^{2}}{x} d x

使用换元积分法

= \int - \frac{u ^{2}}{- u ^{2} + 400} d u

提出常数:

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

= - \int \frac{u ^{2}}{- u ^{2} + 400} d u

使用换元积分法

= - \int \frac{20 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

= - 20 \cdot \int \frac{v ^{2}}{- v ^{2} + 1} d v

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

= - 20 \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

= - 20 (\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

= - 20 (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v)

代回

= - 20 \frac{ln \frac{400 - x ^{2}}{20} + 1}{2} - \frac{ln \frac{400 - x ^{2}}{20} - 1}{2} - \frac{400 - x ^{2}}{20}

化简

- 20 \frac{ln \frac{400 - x ^{2}}{20} + 1}{2} - \frac{ln \frac{400 - x ^{2}}{20} - 1}{2} - \frac{400 - x ^{2}}{20} : - 10 ln \frac{1}{20} - x^{2} + 400 + 1 + 10 ln \frac{1}{20} - x^{2} + 400 - 1 + 400 - x^{2}

= - 10 ln \frac{1}{20} - x^{2} + 400 + 1 + 10 ln \frac{1}{20} - x^{2} + 400 - 1 + 400 - x^{2}

解答补常数

= - 10 ln \frac{1}{20} - x^{2} + 400 + 1 + 10 ln \frac{1}{20} - x^{2} + 400 - 1 + 400 - x^{2} + C

作图

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