Question

∫[log(logx) + {1/(logx)}]dx = _______+ c 

(a) [x/{log(log x)}] 

(b) x + log(logx) 

(c) log(logx) + (1/x) 

(d) xlog(logx)

Answer 1

The correct option (d) xlog(logx)   

Explanation:

I = ∫[log(logx) + {1/(log x)}]dx 

Put logx = t 

∴ x = e^t 

dx = e^tdt 

∴ I = ∫[logt + (1/t)]e^tdt 

This is of the form ∫e^x[f(x) + f'(x)]dx 

whose solution is given by e^xf(x) + c 

∴ by comparing, 

I = e^tlogt + c 

∴ I = xlog(logx) + c