integral of 1/(2xlog_{2)(x)}

解

\int \frac{1}{2 x lo g _{2} ( x )} d x

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

解答ステップ

\int \frac{1}{2 x lo g _{2} ( x )} d x

定数を除く:

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

= \frac{1}{2} \cdot \int \frac{1}{x lo g _{2} ( x )} d x

部分積分を適用する

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

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

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

簡素化

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

定数を解答に追加する

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