依题意有F'(x)=f(x)f(x)F(x)=(arctan√x)/[√x(1+x)]两边分别对x积分,有∫f(x)F(x)dx=∫F(x)dF(x)=1/2*[F(x)]^2=∫(arctan√x)/[√x(1+x)]dx=∫2arctan√x*d(arctan√x)=(arctan√x)^2+C于是得[F(x)]^2=2(arctan√x)...F(x)]^2=2(arctan√x)^2+2C怎么得的？