What is the integral of 1/(xlog_{2)(x)} ?

The integral of 1/(xlog_{2)(x)} is 1/(log_{2)(x)}+1/(ln(2)log_{2)(e)^2}(ln|ln(x)|-1/(ln(x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Simplify

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

Add a constant to the solution

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

\int \frac{1}{( x ^{2} + 1 ) ^{\frac{5}{2}}} d x

\int (ln (sin (x))) d x

\int \frac{1}{y 4 y ^{2} - 1} d y

\int csc^{3} (x) cot^{4} (x) d x

\int 2 x 2 x^{2} + 1 d x

What is the integral of 1/(xlog_{2)(x)} ?
The integral of 1/(xlog_{2)(x)} is 1/(log_{2)(x)}+1/(ln(2)log_{2)(e)^2}(ln|ln(x)|-1/(ln(x)))+C