陈沐涵的回答：

∫dx/√(1＋x^2) let x= tany dx = (secy)^2 dy ∫dx/√(1＋x^2) =∫dy =y+ C =arctanx + C

伤心小男孩的回答：

∫1/[x2√(1+x2)] dx =∫[(x2+1)-x2]/[x2√(1+x2)] dx =∫1/x2dx-∫1/(x2+1)dx =-1/x-arctanx+c