Improper Integrals

Area = ∫ 1/(x^2 + x) dx from x = 1 to x = infinity.
= ∫ [ 1/x - 1/(x + 1) ] dx ( by partial fraction)
= ln x - ln ( x + 1) from x = 1 to x = infinity
(i) lim [ ln x - ln (x + 1)] when x tends to infinity
= lim ln[ x/(x + 1)]
= lim ln [ 1 - 1/(x + 1)]
= ln 1 when x tends to infinity
= 0.
(ii) When x = 1
ln[ x/(x + 1)] = ln (1/2) = - ln 2
so area = (i) - (ii) = 0 - [ - ln 2]
= ln 2.