用换元积分求以下积分∫e^x√(e^x-1)/(e^x+3)dx
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用换元积分求以下积分
∫e^x√(e^x-1)/(e^x+3)dx
∫e^x√(e^x-1)/(e^x+3)dx
![用换元积分求以下积分∫e^x√(e^x-1)/(e^x+3)dx](/uploads/image/z/16064040-48-0.jpg?t=%E7%94%A8%E6%8D%A2%E5%85%83%E7%A7%AF%E5%88%86%E6%B1%82%E4%BB%A5%E4%B8%8B%E7%A7%AF%E5%88%86%E2%88%ABe%5Ex%E2%88%9A%EF%BC%88e%5Ex-1%EF%BC%89%2F%28e%5Ex%2B3%29dx)
令a=√(e^x-1)
e^x=a²+1
x=ln(a²+1)
所以dx=2ada/(a²+1)
x=0,a=0
x=ln5,a=2
所以原式=∫(a²+1)*a/(a²+4)*2ada/(a²+1)
=∫2a²/(a²+4) da
=2∫[1-4/(a²+4)] da
=2∫da-2∫4/(a²+4) da
=2∫da-2∫1/[(a/2)²+1] da
=2∫da-4∫1/[(a/2)²+1] d(a/2)
=2a-4arctan(a/2)
=4-4arctan1
e^x=a²+1
x=ln(a²+1)
所以dx=2ada/(a²+1)
x=0,a=0
x=ln5,a=2
所以原式=∫(a²+1)*a/(a²+4)*2ada/(a²+1)
=∫2a²/(a²+4) da
=2∫[1-4/(a²+4)] da
=2∫da-2∫4/(a²+4) da
=2∫da-2∫1/[(a/2)²+1] da
=2∫da-4∫1/[(a/2)²+1] d(a/2)
=2a-4arctan(a/2)
=4-4arctan1