Question

求∫1／√(x^ 2+1)dx答案，最好有详细过程，我对这些不是很懂

Answer 1

解：设x=tant，则sint=x/√(x²+1)，dx=sec²tdt
∴原式=∫sec²tdt/sect
=∫costdt/cos²t
=∫d(sint)/(1-sin²t)
=(1/2)∫[1/(1+sint)+1/(1-sint)]d(sint)
=(1/2)[ln(1+sint)-ln(1-sint)]+C （C是积分常数）
=(1/2)ln[(1+sint)/(1-sint)]+C
=ln[x+√(x²+1)]+C (把sint=x/√(x²+1)代入，并整理得)。

Answer 2

设x=tant,dx=(sect)^2dt,
sect=√（1+x^2)
原式=∫(sect)^2dt/(sect)
=∫dt/cost
=∫costdt/(cost)^2
=∫d(sint)/[1-(sint)^2]
设sint=u,
原式=∫du/(1-u^2)
=(1/2)∫du/(1+u)+(1/2))∫du/(1-u)
=(1/2)ln|1+u|-(1/2)ln|1-u|+C
=(1/2)ln|(1+sint)/(1-sint)|+C,//分子、分母同乘1+sint
=ln[(1+sint)/cost]+C
=ln|sect+tant|+C
=ln|x+√（1+x^2)|+C.