∫log(log(logx))xlogxlog(logx)du=f(x)+C then f(x)=
12loglogxlogx
12loglogx2
loglog(logx)2
12loglog(logx)2
I=∫log(log(logx))xlogxlog(logx)dx
Let log(log(logx))=tI=1xlogxlog(logx)dx=dtI=∫tdt=t22+C=(log(log(logx)))22+C