Q.273. Indefinite integration.

Question 273.
Find ∫cot-1(1-x+x2)dx.

Answer 273.
This can be done using integration by parts
∫ uv dx = u ∫ vdx - ∫ [(du/dx) ∫ vdx] dx

Taking u = cot^-1 (1 - x + x^2) and v = 1,
∫cot^-1(1 - x + x2) dx
= cot^-1(1 - x + x2) ∫ dx - ∫ [d/dx (cot^-1(1 - x + x^2)) ∫ dx] dx
= x cot^-1 (1 - x + x^2) - ∫ x * [d/dx (cot^-1(1 - x + x^2)) dx ... (1)

cot^-1 (1 - x + x^2)
tan^-1 [1 / (1 - x + x^2)] ... [because 1 - x + x^2 = (x - 1/2)^2 + (√3/2)^2 > 0 for all x.]
= tan^-1 [ [x + (1 - x)] / [1 - x (1 - x)] ]
= tan^-1 x - tan^-1 (x - 1)
=> d/dx [cot^-1 (1 - x + x^2)
= d/dx [tan^-1 x - tan^-1 (x - 1)]
= 1/(1 + x^2) - 1/ (1 + (x - 1)^2)

Plugging this result in (1) above,
Integral
= x cot^-1 (1 - x + x^2) - ∫ [x/(1 + x^2) - x/(x^2 - 2x + 2)] dx
= x cot^-1 (1 - x + x^2) - (1/2) ∫ 2x / (1 + x^2) dx + (1/2) ∫ (2x - 2 + 2) / (x^2 - 2x + 2) dx
= x cot^-1 (1 - x + x^2) - (1/2) ln (1 + x^2) + (1/2) ln (x^2 - 2x + 2) + ∫ dx / [1 + (x - 1)^2]
= x cot^-1 (1 - x + x^2) - (1/2) ln (1 + x^2) + (1/2) ln (x^2 - 2x + 2) + tan^-1 (x - 1) + c.

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