求∫dx/1+(根号x)
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求∫dx/1+(根号x)
解
令√x=t
则t²=x,dx=2tdt
∴∫dx/(1+√x)
=∫2tdt/(t+1)
=2∫[(t+1)-1]/(t+1)dt
=2∫1-1/(t+1)dt
=2t-2ln|t+1|+C
=2√x-2ln(√x+1)+C
再问: =2��1-1/(t+1)dt �����
再答: =2��[1-1/(t+1)]dt����1��ȥ(t+1��֮1)
令√x=t
则t²=x,dx=2tdt
∴∫dx/(1+√x)
=∫2tdt/(t+1)
=2∫[(t+1)-1]/(t+1)dt
=2∫1-1/(t+1)dt
=2t-2ln|t+1|+C
=2√x-2ln(√x+1)+C
再问: =2��1-1/(t+1)dt �����
再答: =2��[1-1/(t+1)]dt����1��ȥ(t+1��֮1)