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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

使用换元积分法

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

提出常数:

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

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \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}

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

代回

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

化简

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

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

解答补常数

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

作图

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