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积分∫e^x(2x+1)/2√x
人气:277 ℃ 时间:2020-06-29 02:23:23
解答
令t=√x,dt=dx/2t;
=∫e^x(2x+1)/(2√x) dx=∫e^(t^2)(2t^2+1) dt=∫e^(t^2) dt+∫2t^2*e^(t^2) dt=∫e^(t^2) dt+∫t*[e^(t^2) d(t^2)]
=∫e^(t^2) dt+[t*e^(t^2)-∫e^(t^2)* dt]=t*e^(t^2)+C
=C+√x*e^x
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