∫²₁(1/x-2的x次方)dx=？
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L = ∫ x/√(x²-x+1) dx
= ∫ x/√[(x-1/2)²+3/4] dx
Let x-1/2 = (√3/2)tany,dx = (√3/2)sec²y dy
tany = (2x-1)/√3,secy=2√(x²-x+1)/√3
L = ∫ [(√3/2)tany + 1/2]secy dy
= (√3/2)∫ secytany dy + (1/2)∫ secy dy
= (√3/2)secy + (1/2)ln| secy+tany | + C
= (√3/2)(2/√3)√(x²-x+1) + (1/2)ln| 2√(x²-x+1)/√3 + (2x-1)/√3 | + C
= √(x²-x+1) + (1/2)ln| (2x-1)+2√(x²-x+1) | + C₁
= ∫ x/√[(x-1/2)²+3/4] dx
Let x-1/2 = (√3/2)tany,dx = (√3/2)sec²y dy
tany = (2x-1)/√3,secy=2√(x²-x+1)/√3
L = ∫ [(√3/2)tany + 1/2]secy dy
= (√3/2)∫ secytany dy + (1/2)∫ secy dy
= (√3/2)secy + (1/2)ln| secy+tany | + C
= (√3/2)(2/√3)√(x²-x+1) + (1/2)ln| 2√(x²-x+1)/√3 + (2x-1)/√3 | + C
= √(x²-x+1) + (1/2)ln| (2x-1)+2√(x²-x+1) | + C₁
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