Question

what is the integral value of logx​

Answer 1

Answer:

Thanx for A2A.

∫logx=xlogx−x+c

Put x=et(i.e.t=logx)

Then dx=etdt

So,

∫logx

=∫log(et)etdt

=∫tet⋅dt

Now using ∫f(t)⋅g(t)dt=f(t)∫g(t)dt−∫f′(t)(∫g(t)dt)dt+c