高数∫1/（a^2-x^2)^2dx详细解答

设x=atant t=arctan(x/a)=1/a*∫cos^2t/cos2t *sec^2tdt=1/2a*∫1/cos2td2t=ln |tan2t+sec2t| + C t=arctan(x/a)--------------------设 u=tant=x/atan2t=2u/1-u^2 sec2t=1/cos2x=1+u^2/1-u^2tan2t+sec2t=(1+u)^2/(1...答案不对呀中间环节出错了。。、=1/a*∫cos^4t/cos^2(2t) *sec^2tdt=(1/4a)*∫(1+cos2t)/cos^2(2t)d2t=(1/4a)*∫1/cos^2(2t)+1/cos2td(2t)=(1/4a)*{tan2t+ln |tan2t+sec2t|} + Ct=arctan(x/a)--------------------设 u=tant=x/atan2t=2u/1-u^2 sec2t=1/cos2x=1+u^2/1-u^2tan2t+sec2t=(1+u)^2/(1-u^2)-------------------=(1/4a)*{2ax/(a^2-x^2)+2ln(1+x/a)-ln(1-(x/a)^2)}+C=(x/2a)/(a^2-x^2)+(1/2a)ln|a+x|-(1/4a)ln|a^2-x^2|+C