integral of ln(1+3x)

解答

\int ln (1 + 3 x) d x

\frac{1}{3} ln (1 + 3 x) + x ln (1 + 3 x) - \frac{1}{3} - x + C

求解步骤

\int ln (1 + 3 x) d x

使用换元积分法

= \int ln (u) \frac{1}{3} d u

提出常数:

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

= \frac{1}{3} \cdot \int ln (u) d u

使用分布积分法

= \frac{1}{3} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{3} (u ln (u) - u)

u = 1 + 3 x

代回

= \frac{1}{3} ((1 + 3 x) ln (1 + 3 x) - (1 + 3 x))

化简

\frac{1}{3} ((1 + 3 x) ln (1 + 3 x) - (1 + 3 x)) : \frac{1}{3} ln (1 + 3 x) + x ln (1 + 3 x) - \frac{1}{3} - x

= \frac{1}{3} ln (1 + 3 x) + x ln (1 + 3 x) - \frac{1}{3} - x

解答补常数

= \frac{1}{3} ln (1 + 3 x) + x ln (1 + 3 x) - \frac{1}{3} - x + C

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