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高数积分定积分求解
1个回答
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设t=x+1
∫1/[(x+1)((x+1)^(5/2)+1)]dx
=∫1/[t(t^(5/2)-1)]dt
=∫t^(3/2)/[t^(5/2)(t^(5/2)-1)]dt
=(2/5)∫[1/(t^(5/2)-1)-1/t^(5/2)]d(t^(5/2))
=(2/5)(ln[(t^(5/2)-1]-ln(t^(5/2))
=(2/5)(ln[((x+1)^(5/2)-1]-ln((x+1)^(5/2))
=(2/5)ln(242*32/(243*31))
=(2/5)ln(7744/7533)
∫1/[(x+1)((x+1)^(5/2)+1)]dx
=∫1/[t(t^(5/2)-1)]dt
=∫t^(3/2)/[t^(5/2)(t^(5/2)-1)]dt
=(2/5)∫[1/(t^(5/2)-1)-1/t^(5/2)]d(t^(5/2))
=(2/5)(ln[(t^(5/2)-1]-ln(t^(5/2))
=(2/5)(ln[((x+1)^(5/2)-1]-ln((x+1)^(5/2))
=(2/5)ln(242*32/(243*31))
=(2/5)ln(7744/7533)
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