令x=(tant)^4
4∫(sint)^2/(cost)^3dt
4∫(sint)^2/(cost)^4d(sint)
令sint=m
4∫m^2/(1+m^2)^2dm
令m=1/n
-4∫1/(n^2-1)^2dn
裂项
-∫[1/(n-1)-1/(n+1)]^2dn
得到
-∫1/(n-1)^2dn-∫1/(n+1)^2dn+2∫1/(n-1)(n+1)dn
积分前面两项最后一项再
1/(n-1)+1/(n+1)+∫1/(n-1)-1/(n+1)dn
1/(n-1)+1/(n+1)+ln(n-1)-ln(n+1)+C
追问比楼上的简便多了哈