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