帮我解一道数学题 已知 y=arcsinh(x)/[根号下（1+x^2)] 求 (1+x^2)dy/dx+xy=1
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y = arcsin(x)/(1+x^2)^(1/2)
dy/dx = {(1+x^2)^(1/2)/(1-x^2)^(1/2) - xarcsin(x)/(1+x^2)^(1/2)}/(1+x^2),
(1+x^2)dy/dx + xy = {(1+x^2)^(1/2)/(1-x^2)^(1/2) - xarcsin(x)/(1+x^2)^(1/2)} + xarcsin(x)/(1+x^2)^(1/2)
= (1+x^2)^(1/2)/(1-x^2)^(1/2)
= (1-x^4)^(1/2)/(1-x^2)
dy/dx = {(1+x^2)^(1/2)/(1-x^2)^(1/2) - xarcsin(x)/(1+x^2)^(1/2)}/(1+x^2),
(1+x^2)dy/dx + xy = {(1+x^2)^(1/2)/(1-x^2)^(1/2) - xarcsin(x)/(1+x^2)^(1/2)} + xarcsin(x)/(1+x^2)^(1/2)
= (1+x^2)^(1/2)/(1-x^2)^(1/2)
= (1-x^4)^(1/2)/(1-x^2)
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