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受欢迎的微积分 >

integral of 1/(y^2-9)

解答

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

解答

- \frac{1}{6} (ln \frac{y}{3} + 1 - ln \frac{y}{3} - 1) + C

求解步骤

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

使用换元积分法

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

因式分解

u^{2} - 1 : - (- u^{2} + 1)

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

u = \frac{y}{3}

代回

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

化简

\frac{1}{3} (- (\frac{ln \frac{y}{3} + 1}{2} - \frac{ln \frac{y}{3} - 1}{2})) : - \frac{1}{6} (ln \frac{y}{3} + 1 - ln \frac{y}{3} - 1)

= - \frac{1}{6} (ln \frac{y}{3} + 1 - ln \frac{y}{3} - 1)

解答补常数

= - \frac{1}{6} (ln \frac{y}{3} + 1 - ln \frac{y}{3} - 1) + C

作图

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