Find the antiderivative of the function y=ln^2x/x^2.

To determine the primitive of te given function, we'll
calculate the indefinite integral of the function.

Int (ln
x)^2dx/x^2

We'll replace ln x by
t.

ln x = t

x =
e^t

We'll differentiate both
sides:

dx/x = dt

Int (ln
x)^2dx/x^2 = Int t^2*e^(-t)dt

We'll solve the integral in
variable t using parts.

Int udv = u*v - Int
vdu

Let u = t^2 => du =
2tdt

dv = e^(-t) dt => v =
-e^(-t)

Int t^2*e^(-t)dt = -t^2*e^(-t) + 2Int
t*e^(-t)dt

We'll integrate again Int t*e^(-t)dt by
parts:

u = t => du =
dt

dv = e^(-t) dt => v =
-e^(-t)

Int t*e^(-t)dt = -t*e^(-t) + Int e^(-t)
dt

Int t*e^(-t)dt = -t*e^(-t) - e^(-t) +
C

Int t^2*e^(-t)dt = -t^2*e^(-t) + 2*[-t*e^(-t) - e^(-t)] +
C

Int t^2*e^(-t)dt = -t^2*e^(-t) - 2t*e^(-t) - 2e^(-t) +
C

Int t^2*e^(-t)dt = -t^2/e^t - 2t/e^t - 2/e^t +
C

Int t^2*e^(-t)dt = -(t^2 + 2t + 2)/e^t +
C

Int (ln x)^2dx/x^2 = -[(ln x)^2 + 2ln x + 2)/x +
C

The antiderivative of the given function is
F(x) = -[(ln x)^2 + 2ln x + 2)/x + C.