If fn(x)=loglog…logx (log is repeated n times)
then ∫xf1(x)f2(x)…fn(x)−1dx is equal to
fn+1(x)+C
fn+1(x)n+1+C
nfn(x)+C
none of these
Put fn(x)=t,dtdx=1xf1(x)…fn−1(x)
So ∫xf1(x)…fn(x)−1dx=∫dtt=logt+C
=fn+1(x)+C.