求不定积分积分1/根号（1+u^2)度du

Question

求不定积分
积分1/根号（1+u^2)度
du

Answer 1

设u=tant,t∈[-π/2,π/2],则du=sectdt,代入原积分,得
∫du/√(u^2+1)
=∫[1/√((tant)^2+1)]*(sect)^2dt
=∫(1/sect)*(sect)^2dt (t∈[-π/2,π/2],sect≥0)
=∫sectdt
=ln|sect+tant|+C
=ln|√(u^2+1)+u|+C