1/(1+x^2)^2dx的不定积分是多少
问题描述：
1/(1+x^2)^2dx的不定积分是多少
答
x=tant,t=arctanx,dx=(sect)^2dt
原积分=S1/(sect)^4 *(sect)^2 dt
=S(cost)2dt
=S(cos2t+1)/2 dt
=1/4*sin2t+t/2+c
=1/4*2x/(x^2+1)+1/2*arctanx+c
=1/2*x/(x^2+1)+1/2*arctanx+c