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受欢迎的微积分 >

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

解答

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

解答

- ln \frac{x}{2} + 1 + ln \frac{x}{2} - 1 + x + C

求解步骤

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

因式分解

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

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

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

\frac{u ^{2}}{- u ^{2} + 1} = \frac{1}{- u ^{2} + 1} - 1

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d u = u

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

u = \frac{x}{2}

代回

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

化简

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

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

解答补常数

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

作图

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