$\int_{1}^{e} \ln \left(x^{2}\right) d x$

Step 1 ：We are given the integral \(\int_{1}^{e} \ln \left(x^{2}\right) d x\).

Step 2 ：We can calculate the integral of a function using the Fundamental Theorem of Calculus, which states that if a function f is continuous over the interval [a, b] and F is an antiderivative of f on [a, b], then the integral from a to b of f(x) dx is equal to F(b) - F(a).

Step 3 ：In this case, the function to be integrated is \(\ln(x^2)\). The antiderivative of this function can be found using the power rule for integration, which states that the integral of x^n dx is \((1/(n+1))x^(n+1)\), and the chain rule for integration, which states that the integral of f(g(x))g'(x) dx is F(g(x)).

Step 4 ：The antiderivative of \(\ln(x^2)\) is \(x*\ln(x^2) - 2x\). We can then evaluate this antiderivative at the limits of integration, e and 1, and subtract the two results to find the value of the integral.

Step 5 ：The value of the integral is 2.0.

Step 6 ：Final Answer: \(\boxed{2}\)