令x=sinu,则:u=arcsinx,dx=cosudu.
∫[(1+x^2)/√(1-x^2)]dx
=∫{[1+(sinu)^2]/√[1-(sinu)^2]}cosudu
=∫[1+(sinu)^2]du
=∫du+∫(sinu)^2du
=u+(1/2)∫(1-cos2u)du
=u+(1/2)∫du-(1/2)∫cos2udu
=u+(1/2)u-(1/4)∫cos2ud(2u)
=(3/2)u-(1/4)sin2u+C
=(3/2)arcsinx-(1/2)sinucosu+C
=(3/2)arcsinx-(1/2)x√(1-x^2)+C
注:若题目不是我所猜测的那样,则请补充说明.