Question

Evaluate : `int 1/("x" [("log x")^2 + 4])  "dx"`

Answer 1

Let I = `int 1/("x" [("log x")^2 + 4])  "dx"`

Put log x = t

Differentiating w.r.t.x 

`1/"x" "dx" = "dt"`

`therefore "I" = int 1/("t"^2 + 4)  "dt"`

`therefore "I" = int 1/("t"^2 + (2)^2) "dt"`

`= 1/2 "tan"^(-1)("t"/2) + c`

By using `int 1/("x"^2 + "a"^2) "dx" = 1/"a" "tan"^(-1) ("x"/"a") + "c"`

∴ I = `1/2 "tan"^(-1) ("log x"/2) + "c"`