求不定积分dx/[(x-1)^4根号(x^2-2x)]=t(2t^2+3)/3(t^2+1)^(3/2)+C 怎么来的啊 麻烦啦
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![求不定积分dx/[(x-1)^4根号(x^2-2x)]=t(2t^2+3)/3(t^2+1)^(3/2)+C 怎么来的啊 麻烦啦](/uploads/image/z/2694192-24-2.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86dx%2F%5B%28x-1%29%5E4%E6%A0%B9%E5%8F%B7%EF%BC%88x%5E2-2x%EF%BC%89%5D%3Dt%282t%5E2%2B3%29%2F3%28t%5E2%2B1%29%5E%283%2F2%29%2BC+%E6%80%8E%E4%B9%88%E6%9D%A5%E7%9A%84%E5%95%8A+++++++%E9%BA%BB%E7%83%A6%E5%95%A6)
求不定积分dx/[(x-1)^4根号(x^2-2x)]=t(2t^2+3)/3(t^2+1)^(3/2)+C 怎么来的啊 麻烦啦
求不定积分dx/[(x-1)^4根号(x^2-2x)]
=t(2t^2+3)/3(t^2+1)^(3/2)+C 怎么来的啊
麻烦啦
求不定积分dx/[(x-1)^4根号(x^2-2x)]=t(2t^2+3)/3(t^2+1)^(3/2)+C 怎么来的啊 麻烦啦
√(x^2-2x)=t,x^2-2x=t^2 (x-1)^2=t^2+1,(x-1)dx=tdt,代入得:
∫dx/[(x-1)^4根号(x^2-2x)]
=∫tdt/t(t^2+1)^(5/2)
=∫dt/(t^2+1)^(5/2)
=t(2t^2+3)/3(t^2+1)^(3/2)+C (查积分表)
最后代t
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