Quote by A Dhingra
Work is W=∫F.dx .
So in it if i apply the limits some distance r to ∞, i would get,
W = ∫(kQq/x^{2}) dx = kQq/x, x from r to ∞,
= kqQ(1/∞1/r) = kQq/r..
this is definitely not ∞.

Exactly! Well done.
Quote by A Dhingra
But I was thinking that the test charge is moving continuously (with non uniform, decreasing acceleration) almost for infinite time. In that case the sum of all the small patches under the curve should have given infinite energy(I was considering kinetic energy, on vt graph)....but that doesn't seem to be happening...

Again, you need to carry out the appropriate integral. However, you seem to think that the area under a vt graph represents energy. It does not.
Think about what the appropriate integrand is, consider impulse, momentum, and the relationship between momentum and KE. Look at that integrand, is there some part of it that makes you think it might not be infinite? If you can, perform the integral.