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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

a^{2} - a^{2} x^{2} = a^{2} 1 - x^{2}

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

提出常数:

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

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

使用换元积分法

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

使用积分常见定则:

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

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

u = 1 - x^{2}

代回

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

化简

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

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

解答补常数

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

作图

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