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受欢迎的微积分 >

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

解答

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

解答

- 8 (- \frac{x ^{3}}{24} - \frac{x}{2} + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1) + C

求解步骤

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

因式分解

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

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

使用换元积分法

= - \int \frac{8 u ^{4}}{- u ^{2} + 1} d u

提出常数:

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

= - 8 \cdot \int \frac{u ^{4}}{- u ^{2} + 1} d u

长除

\frac{u ^{4}}{- u ^{2} + 1} : - u^{2} - 1 + \frac{1}{- u ^{2} + 1}

= - 8 \cdot \int - u^{2} - 1 + \frac{1}{- u ^{2} + 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

= - 8 (- \int u^{2} d u - \int 1 d u + \int \frac{1}{- u ^{2} + 1} d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

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

= - 8 (- \frac{u ^{3}}{3} - u + \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

u = \frac{x}{2}

代回

= - 8 (- \frac{( \frac{x}{2} ) ^{3}}{3} - \frac{x}{2} + \frac{ln \frac{x}{2} + 1}{2} - \frac{ln \frac{x}{2} - 1}{2})

化简

- 8 (- \frac{( \frac{x}{2} ) ^{3}}{3} - \frac{x}{2} + \frac{ln \frac{x}{2} + 1}{2} - \frac{ln \frac{x}{2} - 1}{2}) : - 8 (- \frac{x ^{3}}{24} - \frac{x}{2} + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1)

= - 8 (- \frac{x ^{3}}{24} - \frac{x}{2} + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1)

解答补常数

= - 8 (- \frac{x ^{3}}{24} - \frac{x}{2} + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1) + C

作图

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