1⼀(x+(x^2+X+1)^2)的不定积分

2025-02-23 14:17:51
推荐回答(1个)
回答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