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

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

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

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

arctan (ln (x)) + C

Solution steps

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

Apply u-substitution

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

= \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

\int \frac{sec ^{4} ( x ) tan ( x )}{243} d x

\frac{d}{d x} (sin^{2} (a x))

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

\int \frac{x}{3 x ^{2} + 5} d x

\frac{d}{d x} (\frac{4}{( x + 2 ) ^{2} + 16})

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