求[arctan(1/x)]/[1+(x^2)]的不定积分
问题描述：
求[arctan(1/x)]/[1+(x^2)]的不定积分
答
∫arctan(1/x)/(1+x²)dx
=∫arctanx(1/x)d(arctanx)
=arctan(1/x)arctanx-∫arctanx·(-1/x²)/[1+(1/x)²]dx
=arctan(1/x)arctanx+∫arctanx[1/(1+x²)]dx
=arctan(1/x)arctanx+∫arctanxd(arctanx)
=arctan(1/x)arctanx+(arctanx)²/2+C
答
令t=1/x
原式 = ∫ (arctant)/(1+ 1/t^2) d(1/t)
= - ∫ (arctant) / (t^2 +1) dt
= - ∫ arctant darctant
= -1/2(arctant)^2 +C
= -1/2 ( arctan 1/x )^2 +C