∫x^2*根号1+x^2dx

Question

∫x^2*根号1+x^2 dx

Answer 1

令x=tant,
则原式=∫tan^2(t)/cos^3(t)dt=∫sin^2(t)/cos^5(t)dt=∫sin^2(t)cost/cos^6(t)dt=∫sin^2(t)/(1-sin^2(t)^3d(sint)
剩下的就是有理函数的积分了,比较烦,我就偷懒不做了.