求dx/(1+x+x^2)^(3/2)的不定积分
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∫dx/(x²+x+1)^(3/2)
= ∫dx/[(x+1/2)²+3/4]^(3/2)
令x+1/2=√3/2*tanψ => dx=√3/2*sec²ψ dψ
sinψ=(x+1/2)/√(x²+x+1)，cosψ=(√3/2)/√(x²+x+1)
(x²+x+1)^(3/2)=(3/4*tan²ψ)^(3/2)=(3/4)^(3/2)*(sec²ψ)^(3/2)=(3√3/8)sec³ψ
=> ∫(√3/2*sec²ψ)/(3√3/8 * sec³ψ) dψ
=> (4/3)∫cosψ dψ
=> (4/3)sinψ + C
=> (4/3) * (x+1/2)√(x²+x+1) + C
=> (2/3) * (2x+1)/√(x²+x+1) + C
= ∫dx/[(x+1/2)²+3/4]^(3/2)
令x+1/2=√3/2*tanψ => dx=√3/2*sec²ψ dψ
sinψ=(x+1/2)/√(x²+x+1)，cosψ=(√3/2)/√(x²+x+1)
(x²+x+1)^(3/2)=(3/4*tan²ψ)^(3/2)=(3/4)^(3/2)*(sec²ψ)^(3/2)=(3√3/8)sec³ψ
=> ∫(√3/2*sec²ψ)/(3√3/8 * sec³ψ) dψ
=> (4/3)∫cosψ dψ
=> (4/3)sinψ + C
=> (4/3) * (x+1/2)√(x²+x+1) + C
=> (2/3) * (2x+1)/√(x²+x+1) + C
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