用第一类换元法解比较简便。∫[x/√(9-4x²)]dx=-⅛∫[1/√(9-4x²)]d(9-4x²)=-⅛·2√(9-4x²) +C=-¼√(9-4x²) +C
∫x/√(9-4x^2) dx= -(1/8) ∫d(9-4x^2)/√(9-4x^2) = -(1/4) √(9-4x^2) + C
