Ashwin Muralidharan IIT Madras
Last Activity: 13 Years ago
Hi Suchita,
Here I'm presuming that last term must have been tan(ln2x)
Make the substitution lnx = t,
then 1/xdx = dt.
So the integral reduces to {tant*tan(t-ln2)*tan(t+ln2) dt}
Now tan(t-ln2) = {tant - tan(ln2)}/{1+tant*tan(ln2)}
and tan(t+ln2) = {tant + tan(ln2)}/{1-tant*tan(ln2)}
Substitute it in the integral, and make the substitution tan2t = y
You will have the integral [ (y-a2) / [2{1-(a2y)}(1+y)] ]dy, where a = ln2.
Which is a standard integral of the form, Linear/Quadratic...... which one can easily integrated by the standard method.
Hope that helps,
All the best,
Regards,
Ashwin (IIT Madras).