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

解答

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

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

求解步骤

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

使用分布积分法

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

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

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

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 (x + 1) - 2 (x + 1 - ln ∣ x + 1 ∣)

解答补常数

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

\int \frac{2 x ^{2} + 1}{( x - 3 ) ^{2}} d x

\int \frac{1}{3 5 ^{x}} d x

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

\int \frac{y}{3 - y} d y

\int \frac{1}{x ( 1 + x ) ^{5}} d x