Question

Evaluate:

`int{1/logx - 1/(logx)^2}dx`; (where x > 1).

Answer 1

`int{1/log_ex - 1/(log_ex)^2}dx`

= `intdx/log_ex - int1/(log_ex)^2dx`

= `int1/log_ex int dx - int{d/dx(1/log_ex) intdx}dx - int1/(log_ex)^2dx`

= `x/log_ex + int1/(log_ex)^2 1/x * x*dx - int1/(log_ex)^2dx`

= `x/log_ex + int1/(log_ex)^2dx - intdx/(log_ex)^2`

= `x/log_ex + c`

Where ′c′ is any arbitary constant of integration.