第1个回答 2023-12-27
令x=tanu , u∈(-π/2, π/2)
则dx=sec²udu
原函数=∫1/secu *sec²udu
=∫secudu
=∫1/cosu du
=∫cosu/cos²u du
=∫d(sinu)/(1-sin²u)
=1/2∫d(sinu)[1/(1-sinu)+1/(1+sinu)]
=1/2ln[(1+sinu)/(1-sinu)]+C
=ln|(1+sinu)/cosu|+C
=ln|(1+x/√(x²+1))/(1/√(x²+1))|+C
=ln|(x+√(x²+1)|+C