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是∫dx/((1+3√x)*√x)嗎? 用t=√x,那麼x=t^2,dx=2t ∫(2t/((1+3t)*t))dt=∫(2/(1+3t))dt=(2ln(1+3t))/3+C=(2ln(1+3√x))/3+C 碰到這種有複合函數的可以考慮用換元法,至於是第一類還是第二類就靠你自己判斷了
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