Sum

**Find:**

`int"x".tan^-1 "x" "dx"`

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#### Solution

Let I = ∫x tan^{-1} x dx

Taking tan^{-1} x as first function and x as the second function and integrating by parts, we obtain

`"I" = tan^-1 "x" int"x dx" - int {("d"/"dx"

tan^-1"x")int"x dx"} "dx"`

`= tan^-1"x"("x"^2/2) - int1/(1+"x"^2) . "x"^2/2 "dx"`

` = ("x"^2tan^-1"x")/2 - 1/2 int"x"^2/(1+"x"^2)"dx"`

` = ("x"^2tan^-1"x")/2 - 1/2 int(("x"^2 + 1)/(1+"x"^2) - 1/(1+"x"^2))"dx" `

` = ("x"^2tan^-1"x")/2 - 1/2 int(1 - 1/(1+"x"^2))"dx"`

` = ("x"^2tan^-1"x")/2 - 1/2("x" - tan^-1"x") + "C"`

` = "x"^2/2 tan^-1"x" -"x"/2+ 1/2tan^-1"x"+"C"`

Concept: Comparison Between Differentiation and Integration

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