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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(x+x(ln(x))^2)

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

arctan (ln (x)) + C

Solution steps

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

Simplify

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

Apply u-substitution

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

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= arctan (u)

Substitute back

u = ln (x)

= arctan (ln (x))

Add a constant to the solution

= arctan (ln (x)) + C

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