∫dxx(logx)(loglogx)…(loglog…⏟8 times x) is equal to
(loglog…⏟8 times x)+C
(loglog…⏟7 times x)+C
(loglog…⏟9 times x)+C
None of these
Let I=∫dxx(logx)(loglogx)…(loglog…⏟8 times x)put , (loglog…⏟8 times x)=t
⇒1(loglog…⏟7 times x)…(loglogx)(logx)xdx=dt∴ I=∫dtt=logt+C=(logloglog…⏟9 times x)+C