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Evaluate: ∫ cot–1 (1 + x + x2) dx 

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The given integral is

∫ cot–1 (1 + x + x2) dx

= ∫ tan–1 (1 + x) dx + ∫ tan–1 x dx 

Let 1 + x = t 

dx = dt 

= ∫ tan–1 t dx + ∫ tan–1 x dx

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