What is the integral of 1/(x(ln(x))^2) ?

The integral of 1/(x(ln(x))^2) is -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/(x(ln(x))^2)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = ln (x)

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

Add a constant to the solution

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

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

\int 18 x d x

\frac{d y}{d x}, 8 x^{3} + x^{2} y - x y^{3} = 9

\int \frac{1}{x ^{\frac{4}{5}} ( 1 + x ^{\frac{1}{5}} )} d x

x (\frac{d y}{d x}) = 2 x^{2} + y

What is the integral of 1/(x(ln(x))^2) ?
The integral of 1/(x(ln(x))^2) is -1/(ln(x))+C