Answer
$$ = {{6\,x\,\ln \left(x^2+1\right)-9\,\ln \left(x^2+1\right)-12\,x\, \ln \left(2\,x-3\right)+18\,\ln \left(2\,x-3\right)-164\,x\, \arctan x+246\,\arctan x-182}\over{338\,\left(2\,x-3\right)}} $$
Explanation
This page was created using
Integral Calculator