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受欢迎的微积分 >

integral of 1/(v^2+2v-3)

解答

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

解答

- \frac{1}{4} (ln \frac{v + 1}{2} + 1 - ln \frac{v + 1}{2} - 1) + C

求解步骤

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

因式分解

v^{2} + 2 v - 3 : - (- v^{2} - 2 v + 3)

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

提出常数:

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

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

对

- v^{2} - 2 v + 3

配方:

- (v + 1)^{2} + 4

= - \int \frac{1}{- ( v + 1 ) ^{2} + 4} d v

使用换元积分法

= - \int \frac{1}{- u ^{2} + 4} d u

使用换元积分法

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

提出常数:

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

= - \frac{1}{2} \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}

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

代回

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

化简

- \frac{1}{2} (\frac{ln \frac{v + 1}{2} + 1}{2} - \frac{ln \frac{v + 1}{2} - 1}{2}) : - \frac{1}{4} (ln \frac{v + 1}{2} + 1 - ln \frac{v + 1}{2} - 1)

= - \frac{1}{4} (ln \frac{v + 1}{2} + 1 - ln \frac{v + 1}{2} - 1)

解答补常数

= - \frac{1}{4} (ln \frac{v + 1}{2} + 1 - ln \frac{v + 1}{2} - 1) + C

作图

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