∫2x²+3x-5/x+3dx
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∫2x²+3x-5/x+3dx
∫2x²+3x-5/x+3dx
∫2x²+3x-5/x+3dx
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.