If `l^(prime)(x)` means `logloglog x ,` the `log` being repeated `r` times, then `int[x l(x)l^2(x)l^3(x) l^(prime)(x)]^(-1)d s` is equal to `l^(r+1)(x)+C` (b) `(l^(r+1)(x))/(r+1)+C` `l^r(x)+C` (d) none of these

Options

A. `l^(r+1)(x)+C`
B. `(l^(r+1)(x))/(r+1)+C`
C. `l^(r)(x)+C`
D. non of these

Answer

A

Explanation

Putting `l^(r+1)(x)=t`
and `(1)/(xl(x)l^(2)(x) ...l^(r)(x))dx=dt,` we get
`int(dx)/(xl^(2)(x)l^(3)(x) ...l^(r)(x))=int 1 dt=t+C=l^(r+1)(x)+C`