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受欢迎的微积分 >

integral of (140)/(7xsqrt(100-49x^2))

解答

\int \frac{140}{7 x 100 - 49 x ^{2}} d x

解答

- ln \frac{1}{10} - 49 x^{2} + 100 + 1 + ln \frac{1}{10} - 49 x^{2} + 100 - 1 + C

求解步骤

\int \frac{140}{7 x 100 - 49 x ^{2}} d x

提出常数:

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

= \frac{140}{7} \cdot \int \frac{1}{x 100 - 49 x ^{2}} d x

化简

= 20 \cdot \int \frac{1}{x 100 - 49 x ^{2}} d x

使用换元积分法

= 20 \cdot \int \frac{1}{2 u 100 - 49 u} d u

提出常数:

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

= 20 \cdot \frac{1}{2} \cdot \int \frac{1}{u 100 - 49 u} d u

使用换元积分法

= 20 \cdot \frac{1}{2} \cdot \int - \frac{2}{- v ^{2} + 100} d v

提出常数:

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

= 20 \cdot \frac{1}{2} (- 2 \cdot \int \frac{1}{- v ^{2} + 100} d v)

使用换元积分法

= 20 \cdot \frac{1}{2} (- 2 \cdot \int \frac{1}{10 ( - w ^{2} + 1 )} d w)

提出常数:

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

= 20 \cdot \frac{1}{2} (- 2 \cdot \frac{1}{10} \cdot \int \frac{1}{- w ^{2} + 1} d w)

使用积分常见定则:

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

= 20 \cdot \frac{1}{2} (- 2 \cdot \frac{1}{10} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

代回

= 20 \cdot \frac{1}{2} - 2 \cdot \frac{1}{10} \frac{ln \frac{100 - 49 x ^{2}}{10} + 1}{2} - \frac{ln \frac{100 - 49 x ^{2}}{10} - 1}{2}

化简

20 \cdot \frac{1}{2} - 2 \cdot \frac{1}{10} \frac{ln \frac{100 - 49 x ^{2}}{10} + 1}{2} - \frac{ln \frac{100 - 49 x ^{2}}{10} - 1}{2} : - ln \frac{1}{10} - 49 x^{2} + 100 + 1 + ln \frac{1}{10} - 49 x^{2} + 100 - 1

= - ln \frac{1}{10} - 49 x^{2} + 100 + 1 + ln \frac{1}{10} - 49 x^{2} + 100 - 1

解答补常数

= - ln \frac{1}{10} - 49 x^{2} + 100 + 1 + ln \frac{1}{10} - 49 x^{2} + 100 - 1 + C

作图

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