∫x/(1+√x)dx在0,1定积分
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∫x/(1+√x)dx在0,1定积分
![∫x/(1+√x)dx在0,1定积分](/uploads/image/z/1960046-62-6.jpg?t=%E2%88%ABx%2F%281%2B%E2%88%9Ax%29dx%E5%9C%A80%2C1%E5%AE%9A%E7%A7%AF%E5%88%86)
令t=√x,x=t²,t∈[0,1]
dx=2t
∫[0->1] √x/(1+x)dx=2∫[0->1] t²/(1+t²)dt
=2∫[0->1] 1-[1/(1+t²)]dt
=2-2arctant | [0->1]
=2-π/2
dx=2t
∫[0->1] √x/(1+x)dx=2∫[0->1] t²/(1+t²)dt
=2∫[0->1] 1-[1/(1+t²)]dt
=2-2arctant | [0->1]
=2-π/2