Answer 1

Let I = `int "dx"/(25"x" - "x"(log "x")^2)`

`= int 1/("x"[25 - (log "x")^2])` dx

Put log x = t

∴ `1/"x"` dx = dt

∴ I = `int "dt"/(25 - "t"^2)`

`= int 1/((5)^2 - "t"^2)` dt

`= 1/(2(5)) * log |(5 + "t")/(5 - "t")|` + c

∴ I = `1/10 log |(5 + log "x")/(5 - log "x")|` + c