微积分题目:∫上限5下限1 x/√2x-1 dx
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微积分题目:∫上限5下限1 x/√2x-1 dx
√2x-1=t x=5时,t=3 ; x=1时,t=1
t^2=2x-1
x=(t^2+1)/2
dx=dt
代入原式:
=f (1, 3) (t^2+1)/(2t)dt
=f(1,3) (1/2t+1/(2t))dt
=1/4t^2+1/2lnt (1,3)
=(9/4+1/2ln3)-1/4
=2+1/2ln3
再问: 不对啊 答案是16/3
再答: √2x-1=t x=5时,t=3 ; x=1时,t=1
t^2=2x-1
x=(t^2+1)/2
dx=tdt (这里做错了)
代入原式:
=f (1, 3) [(t^2+1)/(2t)]tdt
=1/2f(1,3) (t^2+1)dt
=1/2 (1/3t^3+t)(1,3)
=1/2 (9+3-1/3-1)
=1/2(12-4/3)
=6-2/3
=16/3
t^2=2x-1
x=(t^2+1)/2
dx=dt
代入原式:
=f (1, 3) (t^2+1)/(2t)dt
=f(1,3) (1/2t+1/(2t))dt
=1/4t^2+1/2lnt (1,3)
=(9/4+1/2ln3)-1/4
=2+1/2ln3
再问: 不对啊 答案是16/3
再答: √2x-1=t x=5时,t=3 ; x=1时,t=1
t^2=2x-1
x=(t^2+1)/2
dx=tdt (这里做错了)
代入原式:
=f (1, 3) [(t^2+1)/(2t)]tdt
=1/2f(1,3) (t^2+1)dt
=1/2 (1/3t^3+t)(1,3)
=1/2 (9+3-1/3-1)
=1/2(12-4/3)
=6-2/3
=16/3
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