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

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

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

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

arctan (ln (x)) + C

Solution steps

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

Apply u-substitution

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

Use the common integral:

\int \frac{1}{u ^{2} + 1} 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(ln^2(x)+1)) ?
The integral of 1/(x(ln^2(x)+1)) is arctan(ln(x))+C