Question

If `int [log(log x) + 1/(logx)^2]dx` = x [f(x) – g(x)] + C, then ______.

Options

• f(x) = log (log x); g(x) = `1/logx`

• f(x) = log x; g(x) = `1/logx`

• f(x) = `1/logx`; g(x) = log (log x)

• f(x) = `1/(xlogx)`; g(x) = `1/logx`

Answer 1

If `int [log(log x) + 1/(logx)^2]dx` = x [f(x) – g(x)] + C, then `underlinebb(f(x) = log (log x); g(x) = 1/logx)`.

Explanation:

Given, 

`int[log(logx) + 1/(logx)^2]dx` = x [f(x) – g(x)] + C

LHS = \[\ce{\int \underset{II}{1} . log (\underset{I}{l}og x) dx + \int \frac{1}{(log x)^2} dx}\]

= `x log (log x) - int 1/logx dx + int 1/(log x)^2 dx`

= `x log (log x) - x/logx - int 1/(logx)^2 dx + int 1/(logx)^2 dx + C`

= `x[log(log x) - 1/logx] + C`

∴ f(x) = log (log x); g(x) = `1/logx`