integral of 1/(-x^2-a+b)

解答

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

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

求解步骤

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

使用换元积分法

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

提出常数:

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

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

u = \frac{x}{- a + b}

代回

= \frac{1}{- a + b} \frac{ln \frac{x}{- a + b} + 1}{2} - \frac{ln \frac{x}{- a + b} - 1}{2}

化简

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

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

解答补常数

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

x \to 1 lim (x^{\frac{1}{x ^{2} - 1}})

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