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

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### 1 Answer

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.**