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integral of (5x)/(x^4-16)

解答

\int \frac{5 x}{x ^{4} - 16} d x

解答

- \frac{5}{16} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1) + C

求解步骤

\int \frac{5 x}{x ^{4} - 16} d x

提出常数:

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

= 5 \cdot \int \frac{x}{x ^{4} - 16} d x

使用换元积分法

= 5 \cdot \int \frac{1}{2 ( u ^{2} - 16 )} d u

提出常数:

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} - 16} d u

使用换元积分法

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{4 ( v ^{2} - 1 )} d v

提出常数:

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

= 5 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2} - 1} d v

因式分解

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

= 5 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

提出常数:

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

= 5 \cdot \frac{1}{2} \cdot \frac{1}{4} (- \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}

= 5 \cdot \frac{1}{2} \cdot \frac{1}{4} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

代回

= 5 \cdot \frac{1}{2} \cdot \frac{1}{4} - \frac{ln \frac{x ^{2}}{4} + 1}{2} - \frac{ln \frac{x ^{2}}{4} - 1}{2}

化简

5 \cdot \frac{1}{2} \cdot \frac{1}{4} - \frac{ln \frac{x ^{2}}{4} + 1}{2} - \frac{ln \frac{x ^{2}}{4} - 1}{2} : - \frac{5}{16} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1)

= - \frac{5}{16} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1)

解答补常数

= - \frac{5}{16} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1) + C

作图

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