在原式里令arctanx=t
则f(t)=tant(1+tan^2(t))^5=sint/cost*1/cos^10(t)=sint/cos^11(t)
所以∫f(x)dx=∫sinx/cos^11(x)dx=-∫1/cos^11(x)d(cosx)=1/(10cos^10(x))+C