What is the integral of (arctan(ln(x)))/x ?

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

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integral of (arctan(ln(x)))/x

\int \frac{arctan ( ln ( x ) )}{x} d x

ln (x) arctan (ln (x)) - \frac{1}{2} ln ln^{2} (x) + 1 + C

Solution steps

\int \frac{arctan ( ln ( x ) )}{x} d x

Apply u-substitution

: \int arctan (u) d u

= \int arctan (u) d u

Apply Integration By Parts

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

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

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

= u arctan (u) - \frac{1}{2} ln u^{2} + 1

Substitute back

u = ln (x)

= ln (x) arctan (ln (x)) - \frac{1}{2} ln ln^{2} (x) + 1

Add a constant to the solution

= ln (x) arctan (ln (x)) - \frac{1}{2} ln ln^{2} (x) + 1 + C

f (x) = x^{2} sin (x^{2} + 3)

(\frac{7 y ^{2} - t ^{2}}{y ^{5}}) \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

\int \frac{x ^{2} - 9}{x ^{2}} d x

y^{^{''}} - (\frac{6}{x}) y^{^{'}} - (\frac{4}{x ^{2}}) y = 0

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