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

The integral of (ln(x))/((1+x^2)) is ln(x)arctan(x)-1/2 i(\Li_{2}(-ix)-\Li_{2}(ix))+C

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

integral

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

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

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{arctan ( x )}{x} d x = \frac{1}{2} i (Li_{2} (- i x) - Li_{2} (i x))

= ln (x) arctan (x) - \frac{1}{2} i (Li_{2} (- i x) - Li_{2} (i x))

Add a constant to the solution

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

\frac{( x + 5 ) ^{3}}{2}

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

\frac{10 t}{( t + 1 ) ^{2}}

8 + x

\frac{( a ^{x} )}{( a ^{x} - 1 )}

What is the integral of (ln(x))/((1+x^2)) ?
The integral of (ln(x))/((1+x^2)) is ln(x)arctan(x)-1/2 i(\Li_{2}(-ix)-\Li_{2}(ix))+C