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受欢迎的微积分 >

integral of 1/(x^2+2x-4)

解答

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

解答

- \frac{1}{2 5} (ln \frac{x + 1}{5} + 1 - ln \frac{x + 1}{5} - 1) + C

求解步骤

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

因式分解

x^{2} + 2 x - 4 : - (- x^{2} - 2 x + 4)

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

提出常数:

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

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

对

- x^{2} - 2 x + 4

配方:

- (x + 1)^{2} + 5

= - \int \frac{1}{- ( x + 1 ) ^{2} + 5} d x

使用换元积分法

= - \int \frac{1}{- u ^{2} + 5} d u

使用换元积分法

= - \int \frac{1}{5 ( - v ^{2} + 1 )} d v

提出常数:

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

= - \frac{1}{5} \cdot \int \frac{1}{- v ^{2} + 1} d v

使用积分常见定则:

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

= - \frac{1}{5} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

代回

= - \frac{1}{5} \frac{ln \frac{x + 1}{5} + 1}{2} - \frac{ln \frac{x + 1}{5} - 1}{2}

化简

- \frac{1}{5} \frac{ln \frac{x + 1}{5} + 1}{2} - \frac{ln \frac{x + 1}{5} - 1}{2} : - \frac{1}{2 5} (ln \frac{x + 1}{5} + 1 - ln \frac{x + 1}{5} - 1)

= - \frac{1}{2 5} (ln \frac{x + 1}{5} + 1 - ln \frac{x + 1}{5} - 1)

解答补常数

= - \frac{1}{2 5} (ln \frac{x + 1}{5} + 1 - ln \frac{x + 1}{5} - 1) + C

作图

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