求∫e∧3√xdx的不定积分
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解：
令√x=t，则x=t^2，dx=2tdt
原积分
=∫(e^3t)2tdt
=(2/3)∫tde^(3t)
=(2/3)[te^(3t)-(1/3)∫e^(3t)d(3t)]
=(2/3)[te^(3t)-(1/3)∫de^(3t)]
=(2/3)[te^(3t)-(1/3)e^(3t)]
=(2/3)e^(3t)(t-1/3)
令√x=t，则x=t^2，dx=2tdt
原积分
=∫(e^3t)2tdt
=(2/3)∫tde^(3t)
=(2/3)[te^(3t)-(1/3)∫e^(3t)d(3t)]
=(2/3)[te^(3t)-(1/3)∫de^(3t)]
=(2/3)[te^(3t)-(1/3)e^(3t)]
=(2/3)e^(3t)(t-1/3)
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令x=t²，dx=2tdt
∫(e^3√x)dx
=∫(e^3t)2tdt
=(2/3)∫tde^(3t)
=(2/3)[te^(3t)-∫e^(3t)dt]
=(2/3)[te^(3t)-(1/3)∫e^(3t)d(3t)]
=(2/3)[te^(3t)-(1/3)∫de^(3t)]
=(2/3)[te^(3t)-(1/3)e^(3t)]
=(2/3)e^(3t)(t-1/3)
=(2/3)e^(3√x)(√x-1/3)
∫(e^3√x)dx
=∫(e^3t)2tdt
=(2/3)∫tde^(3t)
=(2/3)[te^(3t)-∫e^(3t)dt]
=(2/3)[te^(3t)-(1/3)∫e^(3t)d(3t)]
=(2/3)[te^(3t)-(1/3)∫de^(3t)]
=(2/3)[te^(3t)-(1/3)e^(3t)]
=(2/3)e^(3t)(t-1/3)
=(2/3)e^(3√x)(√x-1/3)
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