Q.369. Integration.

Question 369.
Evaluate  ∫ dx / (x^2 + 2x + 2)^2.

Answer 369.
x^2 + 2x + 2
= (x + 1)^2 + 1

Let x + 1 = tanu
=> (x + 1)^2 + 1 = tan^2 u + 1 = sec^2 u
and dx = sec^2 u du

=> Integral
= ∫ sec^2 u du / sec^4 u
= (1/2) ∫ 2cos^2 u du
= (1/2) ∫ (1 + cos2u) du
= (1/2) u + (1/4)sin2u + c
= (1/2) u + (1/2) tanu / (1 + tan^2 u) + c
= (1/2) arctan (x + 1) + (1/2) (x + 1) / [(x + 1)^2 + 1] + c
= (1/2) arctan (x + 1) + (1/2)(x + 1)/(x^2 + 2x + 2) + c
= (1/2) [arctan(x + 1) + (x + 1)/(x^2 + 2x + 2)] + c.

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