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

解答

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

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

求解步骤

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

提出常数:

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

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

使用分布积分法

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

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

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

化简

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

解答补常数

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

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n = 1 \sum \infty \frac{1 0 ^{n}}{9 ^{n}}

x \frac{d y}{d x} - y = x^{2} e^{x}

\frac{\partial}{\partial x} (ln (3 x + 5 y))

x \to π lim ((2 x - 2 π) cot (x))