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

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

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

integral

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

t^{- 2}

e^{- 3 x}

(3 - x)^{3}

(2 x + 5)^{- 1}

\frac{tan ^{2} ( x )}{2}

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