Question

判断下列无穷积分的敛散性,若收敛,则求其值 ∫0 +∞ dx/ [(x+1)√(x^2+1)]

Answer 1

当x趋于无穷时,被积函数等价于1/x^2,因此积分收敛.

做变量替换x=tant,dx=sec^2tdt,x=0对应t=0,x趋于无穷对应t趋于pi/2,

因此原积分=

积分(从0到pi/2)sec^2tdt/[(1+tant)*sect]

=积分(从0到pi/2)dt/(sint+cost)

=1/根号(2) *ln(sin(x+pi/4)/(1+cos(x+pi/4)))|上限pi/2下限0

=1/根号(2)*ln[(2+根号(2))/(2-根号(2))]

=1/根号(2)*ln(3+2根号(2))

=根号(2)*ln(根号(2)+1)