1/(1+x^2)^2dx的不定积分是多少

Question

如题所述

Answer 1

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本回答被提问者采纳

Answer 2

x=tant t=arctanx
1/(1+x^2)^2=(cost)^4
dx/dt=1/cos^2t dx=dt/cos^2t
∫1/(1+x^2)^2dx=∫(cost)^2dt=1/2∫(cos2t+1)dt=1/4sin2t+1/2t+c
=1/4sin(2arctanx)+1/2*(arctanx)+c