Question

1/(1+x^2)^2的不定积分

Answer 1

令x = tanθ，dx = sec²θdθ
∫ dx/(1 + x²)² = ∫ 1/(1 + tan²θ)² · sec²θdθ
= ∫ 1/sec⁴θ · sec²θdθ
= ∫ cos²θdθ
= (1/2)∫ (1 + cos2θ)dθ
= (1/2)(θ + 1/2 · sin2θ) + C
= θ/2 + (1/2)sinθcosθ + C
= (1/2)arctan(x) + (1/2)(x/√(1 + x²))(1/√(1 + x²)) + C
= (1/2)arctan(x) + x/[2(1 + x²)] + C

Answer 2

I=∫ 1/(1+x^2)^2 dx
let
x= tana
dx =(seca)^2 da
I= ∫ [1/(seca)^4] (seca)^2 da
= ∫(cosa)^2 da
= ∫(cos2a+1)/2 da
= [sin(2a)]/4 + a/2 + C
=x/[2(1+x^2)] + (arctanx)/2 +C