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Question
Evaluate the following : `int x tan^-1 x .dx`
Solution
Let I = `int x tan^-1 x .dx`
= `int (tan^-1 x)x.dx`
= `(tan^-1 x) int x.dx - int[{d/dx(tan^-1 x) intx.dx}].dx`
= `(tan^-1x) (x^2/2) - int (1/(1 + x^2)) (x^2/2).dx`
= `(x^2 tan^-1)/(2) - (1)/(2) int x^2/(x^2 + 1).dx`
= `x^2/(2) tan^-1x - (1)/(2) ((x^2 + 1)-1)/(x^2 + 1).dx`
= `x^2/(2)tan^-1x - (1)/(2)[int(1 - 1/(x^2 + 1)).dx]`
= `x^2/(2)tan^-1x - (1)/(2)[int 1.dx - int(1)/(x^2 + 1).dx]`
= `x^2/(2)tan^-1 x - (1)/(2)(x - tan^-1x) + c`.
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