First, let's solve the indefinite integral with trig substitution:
#int1/(x^2+2)^(3/2)dx = int1/(2sqrt(2)((x/sqrt(2))^2+1)^(3/2))dx#
Let #x/sqrt(2) = tan(theta) => 1/sqrt(2)dx = sec^2(theta)d theta#
#=>int1/(2sqrt(2)((x/sqrt(2))^2+1)^(3/2))dx = 1/2intsec^2(theta)/(tan^2(theta)+1)^(3/2)d theta#
#=1/2intsec^2(theta)/sec^3(theta)d theta#
#=1/2intcos(theta)d theta#
#=1/2sin(theta) + C#
#=x/(2sqrt(x^2+2)) + C#
(To do the last step, try drawing the right triangle with an angle #theta# such that #tan(theta) = x/sqrt(2)# and then evaluate #sin(theta)#)
Now, we can look at the definite integral from #-oo# to #oo#
#int_(-oo)^(oo)1/(x^2+2)^(3/2) = lim_(A->-oo)lim_(B->oo)int_A^B1/(x^2+2)^(3/2)dx#
#=lim_(A->-oo)lim_(B->oo)[x/(2sqrt(x^2+2))]_A^B#
#=lim_(A->-oo)lim_(B->oo)1/2(B/sqrt(B^2+2)-A/sqrt(A^2+2))#
#=lim_(A->-oo)lim_(B->oo)1/2(1/(1/Bsqrt(B^2+2))-1/(1/Asqrt(A^2+2)))#
#=lim_(A->-oo)lim_(B->oo)1/2(1/sqrt((B^2+2)/B^2)+1/sqrt((A^2+2)/A^2))#
(We actually cheated a little bit on this step, as we implicitly used that #1/B = 1/sqrt(B^2)# and #1/A = -1/sqrt(A^2)#, which is true given that #B# will be positive and #A# will be negative as we take the limit, but isn't strictly valid to say at this step. It does, however, save us some messy algebra/calculus, and we get the same result)
#= lim_(A->-oo)lim_(B->oo)1/2(1/sqrt(1+2/B^2) + 1/sqrt(1+2/A^2))#
#= lim_(A->-oo)1/2(1 + 1/sqrt(1+2/A^2))#
#=1/2(1+1)#
#=1#
