Question

大一题目，求反常积分？

第四题求反常积分

Answer 1

(4)原式=lnx/[2(1-x^2）]|<2,+∞>-∫<2,+∞>dx/[2x(1-x^2）]
=(1/6)ln2-(1/2)∫<2,+∞>[1/x+x/(1-x^2)]dx
=(1/6)ln2-(1/2)[lnx-(1/2)ln|1-x^2|]|<2,+∞>
=(1/6)ln2+(1/2)[ln2-(1/2)ln3]
=(2/3)ln2-(1/4)ln3.
注：x-->+∞时x/√|1-x^2|-->1.

Answer 2

原式=∫(2,+∞) lnx*x/[(x+1)^2*(x-1)^2]dx
=(1/4)*∫(2,+∞) lnx*[1/(x-1)^2-1/(x+1)^2]dx
=(1/4)*∫(2,+∞) lnx/(x-1)^2dx-(1/4)*∫(2,+∞) lnx/(x+1)^2dx
=(-1/4)*∫(2,+∞) lnxd[1/(x-1)]+(1/4)*∫(2,+∞) lnxd[1/(x+1)]
=(-1/4)*lnx/(x-1)|(2,+∞)+(1/4)*∫(2,+∞) dx/x(x-1)+(1/4)*lnx/(x+1)|(2,+∞)-(1/4)*∫(2,+∞) dx/x(x+1)
=(1/4)*ln2+(1/4)*∫(2,+∞) [1/(x-1)-1/x]dx-(1/12)*ln2-(1/4)*∫(2,+∞) [1/x-1/(x+1)]dx
=(1/6)*ln2+(1/4)*ln|(x-1)/x||(2,+∞)-(1/4)*ln|x/(x+1)||(2,+∞)
=(1/6)*ln2+(1/4)*ln2-(1/4)*ln3+(1/4)*ln2
=(2/3)*ln2-(1/4)*ln3