已知ln[x+√(1+x^2)]是f(x)的原函数
则，f(x)={ln[x+√(1+x^2)]}'
={1/[x+√(1+x^2)]*[x+√(1+x^2)]'
={1/[x+√(1+x^2)]}*{1+(1/2)*[1/√(1+x^2)]*2x}
={1/[x+√(1+x^2)]}*{1+[x/√(1+x^2)]}
=1/√(1+x^2)
∫xf'(x)dx=∫xd[f(x)]=x*f(x)-∫f(x)dx
=x*[1/√(1+x^2)]-ln[x+√(1+x^2)]+C
=[x/√(1+x^2)]-ln[x+√(1+x^2)]+C.
答：因为(arctanx)' = 1/(1+x^2) 所以(arctan(1/x))' = {1/[1+(1/x)^2]}*(1/x)' = - 1/(1+x^2)...详情>>
答：详情>>
