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受欢迎的微积分 >

integral of (ln(3x+1))/(6x+2)

解答

\int \frac{ln ( 3 x + 1 )}{6 x + 2} d x

解答

\frac{1}{12} ln^{2} (3 x + 1) + C

求解步骤

\int \frac{ln ( 3 x + 1 )}{6 x + 2} d x

\frac{ln ( 3 x + 1 )}{6 x + 2} = \frac{1}{2} \frac{ln ( 3 x + 1 )}{3 x + 1}

= \int \frac{1}{2} \cdot \frac{ln ( 3 x + 1 )}{3 x + 1} 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 \frac{ln ( 3 x + 1 )}{3 x + 1} d x

使用换元积分法

= \frac{1}{2} \cdot \int \frac{ln ( u )}{3 u} d u

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{3} \cdot \int \frac{ln ( u )}{u} d u

使用换元积分法

= \frac{1}{2} \cdot \frac{1}{3} \cdot \int v d v

使用幂法则

= \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{v ^{2}}{2}

代回

= \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{ln ^{2} ( 3 x + 1 )}{2}

化简

\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{ln ^{2} ( 3 x + 1 )}{2} : \frac{1}{12} ln^{2} (3 x + 1)

= \frac{1}{12} ln^{2} (3 x + 1)

解答补常数

= \frac{1}{12} ln^{2} (3 x + 1) + C

作图

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