Question

4.求不定积分: (x^2+1)/(x^2+2x+2)^2dx

Answer 1

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Answer 2

(x^2+1)/(x^2+2x+2)^2 = (x^2+2x+2-2x-1)/(x^2+2x+2)^2
= 1/(x^2+2x+2) - (2x+1)/(x^2+2x+2)^2
= 1/(x^2+2x+2) - (2x+2-1)/(x^2+2x+2)^2
= 1/[(x+1)^2+1] - (2x+2)/(x^2+2x+2)^2 + 1/(x^2+2x+2)^2
I = ∫(x^2+1)/(x^2+2x+2)^2dx

= ∫(1/[(x+1)^2+1]dx - ∫[1/(x^2+2x+2)^2]d(x^2+2x+2)
+ ∫{1/[(x+1)^2+1]^2}dx
= arctan(x+1) + 1/(x^2+2x+2) + ∫{1/[(x+1)^2+1]^2}dx
令 x +1 = tant, 则 dx = (sect)^2dt
∫{1/[(x+1)^2+1]^2}dx = ∫(sect)^2dt/(sect)^4 = ∫(cost)^2dt
= (1/2)∫(1+cos2t)dt = t/2 + (1/4)sin2t
= (1/2)arctan(x+1) + (1/2)(x+1)/(x^2+2x+2)
得 I = arctan(x+1) + 1/(x^2+2x+2) + (1/2)arctan(x+1) + (1/2)(x+1)/(x^2+2x+2) + C
= (3/2)arctan(x+1) + (1/2)(x+3)/(x^2+2x+2) + C