∫(arctan√x)/[√x*(1+x)]dx

证明：令t=√x =>x=t^2 dx=2*t*dt
∫(arctan√x)/[√x*(1+x)]dx=∫(arctan(t)/(t*(1+t^2))*2*t*dt
=2∫(arctan(t)/(1+t^2)*dt
=2∫arctan(t)*d(arctan(t))
=arctan^2(t)+C
证毕！

d(artan√x) = 1/(1+x) d(√x) = 1/2√x*(1+x) dx;
∫(arctan√x)/[√x*(1+x)]dx
= ∫ 2(arctan√x) d(arctan√x)
= (arctan√x)^2 + C .

一步一步微分、积分并用,就可以还原出原函数,也就是一些教师所说的“还原法”,或“凑微分法”：∫(arctan√x)/[√x×(1+x)]dx=2∫(arctan√x)/[1+x]d√x=2∫(arctan√x)/[1+(√x)²]d√x=2∫(arctan√x)d(arctan...