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

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

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

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

Solution steps

\int \frac{1}{x ( ln ( x ) + 1 ) ^{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) + 1

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

Add a constant to the solution

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

\int 3 a x \cdot e^{- 2 b x} d x

\int (2 x - 1)^{2} - 9 d x

\int e^{x} 13 + e^{x} d x

\int \frac{- cos ( x )}{sin ^{9} ( x )} d x

\int x cos (- 3 x) d x

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