令x=tanθ,dx=(secθ)^2dθ,∫√（x∧2＋1）dx=∫（secθ)^3dθ=∫[secθ+(tanθ)^2*secθ]dθ=∫secθdθ+∫tanθd(secθ)=∫secθdθ+tanθ*(secθ)-∫(secθ)^3dθ,故,∫√（x∧2＋1）dx=∫（secθ)^3dθ=1/2【∫secθdθ+tanθ*(secθ)】=1/2【ln｜secx+tanx｜+secxtanx]=1/2[ln｜√（x∧2＋1）+x｜+x√（x∧2＋1）]