高数求极限。过程。。。。。
2个回答
展开全部
lim(x->+∞) [∫(0->x) [( t^2-1)/(t^2+1) ] ^2 dt -x ]
=lim(x->+∞) [x-2arctanx +2x/(1+x^2) -x ]
=lim(x->+∞) [-2arctanx + 2x/(1+x^2) ]
=-π
//
∫ [( t^2-1)/(t^2+1) ] ^2 dt
= ∫ [1 - 2/(t^2+1) ] ^2 dt
=∫[ 1 - 4/(t^2+1) + 4/(t^2+1)^2 ] dt
= t - 4arctanu + 4∫dt/(t^2+1)^2
= t - 4arctanu + 2[ arctant +t/(1+t^2) ] +C
= t - 2arctanu + 2t/(1+t^2) +C
∫(0->x) [( t^2-1)/(t^2+1) ] ^2 dt
=[t - 2arctanu + 2t/(1+t^2)] |(0->x)
= x-2arctanx +2x/(1+x^2)
//
let
t=tanu
dt = (secu)^2 du
∫dt/(t^2+1)^2
=∫(cosu)^2 du
=(1/2)∫(1+cos2u) du
=(1/2)[ u +(1/2)sinu ] +C'
=(1/2)[ arctant +t/(1+t^2) ] +C'
//
=lim(x->+∞) [x-2arctanx +2x/(1+x^2) -x ]
=lim(x->+∞) [-2arctanx + 2x/(1+x^2) ]
=-π
//
∫ [( t^2-1)/(t^2+1) ] ^2 dt
= ∫ [1 - 2/(t^2+1) ] ^2 dt
=∫[ 1 - 4/(t^2+1) + 4/(t^2+1)^2 ] dt
= t - 4arctanu + 4∫dt/(t^2+1)^2
= t - 4arctanu + 2[ arctant +t/(1+t^2) ] +C
= t - 2arctanu + 2t/(1+t^2) +C
∫(0->x) [( t^2-1)/(t^2+1) ] ^2 dt
=[t - 2arctanu + 2t/(1+t^2)] |(0->x)
= x-2arctanx +2x/(1+x^2)
//
let
t=tanu
dt = (secu)^2 du
∫dt/(t^2+1)^2
=∫(cosu)^2 du
=(1/2)∫(1+cos2u) du
=(1/2)[ u +(1/2)sinu ] +C'
=(1/2)[ arctant +t/(1+t^2) ] +C'
//
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询