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受欢迎的微积分 >

integral of 2/(4x^2-1)

解答

\int \frac{2}{4 x ^{2} - 1} d x

解答

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

求解步骤

\int \frac{2}{4 x ^{2} - 1} d x

提出常数:

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

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

因式分解

4 x^{2} - 1 : - (- 4 x^{2} + 1)

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

提出常数:

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

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

使用换元积分法

= 2 (- \int \frac{1}{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 (- \frac{1}{2} \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}

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

u = 2 x

代回

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

化简

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

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

解答补常数

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

作图

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