integral of ln((1+x)^2)

解答

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

2 x ln (1 + x) - 2 (1 + x - ln ∣ 1 + x ∣) + C

求解步骤

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

使用分布积分法

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

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

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

使用对数运算法则:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

假定

x \geq 0

= 2 x ln (1 + x) - 2 (1 + x - ln ∣ 1 + x ∣)

解答补常数

= 2 x ln (1 + x) - 2 (1 + x - ln ∣ 1 + x ∣) + C

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