Question

根号下(1+X平方)的原函数

Answer 1

令x=tan(t),t∈(-pi/2,pi/2),则根号(1+x^2)=sec(t),
∫根号(1+x^2)dx
=∫sec(t)d(tan(t))-----(令此积分为I)
=tan(t)sec(t)-∫tan(t)d(sec(t))
=tan(t)sec(t)-∫tan(t)^2.sec(t)dt
=tan(t)sec(t)-∫sec(t)[sec(t)^2-1]dt
=tan(t)sec(t)-∫sec(t)d(tan(t))+∫sec(t)dt
=tan(t)sec(t)-∫sec(t)d(tan(t))+ln[sec(t)+tan(t)]
=tan(t)sec(t)+ln[sec(t)+tan(t)]-I
所以2I=tan(t)sec(t)+ln[sec(t)+tan(t)]+C
I={tan(t)sec(t)+ln[sec(t)+tan(t)]}/2+C
={x根号(1+x^2)+ln[根号(1+x^2)+x]}/2+C
不定积分I即为所求原函数.