solve the integral of 1/ (t^2 - sqrt(2) t+ (1/2)) using completing the square

Question

I already asked the question above

Answer 1

∫1/(t^2 + √2.t +1/2) dt

∫1/{(t - √2/2)^2 + 1/2 -1/2} dt

=∫1/(t -√2/2)^2 dt

let....... u = t - √2/2............ du/dx = 1.................... du = dx

=∫1/u^2 du

= -u^-1 + c

∫1/(t^2 + √2.t +1/2) dt= -(t -√2/2)^-1 + c