设x+3＝t→dx＝dt,代入原式得
∫[(2x²+3x-5)/(x+3)]dx
＝∫[(2(t-3)²+3(t-3)-5)/t]dt
＝∫[2t+(4/t)-9]dt
＝t²+4lnt-9t+C
＝(x+3)²+4ln(x+3)-9(x+3)+C
＝x²-3x-18+4ln(x+3)+C.