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

解答

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

解答

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

求解步骤

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

因式分解

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

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

对

- x^{2} - 2 x + 1

配方:

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

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

使用换元积分法

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

使用换元积分法

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

代回

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

化简

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

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

解答补常数

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

作图

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