integral of 1/(6xlog_{7)(x)}

解答

\int \frac{1}{6 x lo g _{7} ( x )} d x

\frac{1}{6} (\frac{1 - x lo g _{7} ( x )}{lo g _{7} ( x )} + x - \frac{1}{ln ( 7 ) lo g _{7} ( e ) ^{2}} (- ln ∣ ln (x) ∣ + \frac{1}{ln ( x )})) + C

求解步骤

\int \frac{1}{6 x lo g _{7} ( x )} d x

提出常数:

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

= \frac{1}{6} \cdot \int \frac{1}{x lo g _{7} ( x )} d x

\frac{1}{x lo g _{7} ( x )} = \frac{1 - x lo g _{7} ( x )}{x lo g _{7} ( x )} + 1

= \frac{1}{6} \cdot \int \frac{1 - x lo g _{7} ( x )}{x lo g _{7} ( x )} + 1 d x

使用分布积分法

= \frac{1}{6} (\frac{1 - x lo g _{7} ( x )}{lo g _{7} ( x )} + x - \int \frac{- ln ( x ) - 1}{ln ( 7 ) x lo g _{7} ( x ) ^{2}} d x)

\int \frac{- ln ( x ) - 1}{ln ( 7 ) x lo g _{7} ( x ) ^{2}} d x = \frac{1}{ln ( 7 ) lo g _{7} ( e ) ^{2}} (- ln ∣ ln (x) ∣ + \frac{1}{ln ( x )})

= \frac{1}{6} (\frac{1 - x lo g _{7} ( x )}{lo g _{7} ( x )} + x - \frac{1}{ln ( 7 ) lo g _{7} ( e ) ^{2}} (- ln ∣ ln (x) ∣ + \frac{1}{ln ( x )}))

解答补常数

= \frac{1}{6} (\frac{1 - x lo g _{7} ( x )}{lo g _{7} ( x )} + x - \frac{1}{ln ( 7 ) lo g _{7} ( e ) ^{2}} (- ln ∣ ln (x) ∣ + \frac{1}{ln ( x )})) + C

\int 3 x^{2} y + e^{y} d x

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