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Solution
Your input: calculate $$$\int_{2}^{4}\left( - \ln{\left(x^{2} \right)}^{2} + \ln{\left(x^{2} \right)} \right)dx=\int_{2}^{4}\left( - 4 \ln{\left(x \right)}^{2} + 2 \ln{\left(x \right)} \right)dx$$$
First, calculate the corresponding indefinite integral: $$$\int{\left(- 4 \ln{\left(x \right)}^{2} + 2 \ln{\left(x \right)}\right)d x}=2 x \left(- 2 \ln{\left(x \right)}^{2} + 5 \ln{\left(x \right)} - 5\right)$$$ (for steps, see indefinite integral calculator)
According to the Fundamental Theorem of Calculus, $$$\int_a^b F(x) dx=f(b)-f(a)$$$, so just evaluate the integral at the endpoints, and that's the answer.
$$$\left(2 x \left(- 2 \ln{\left(x \right)}^{2} + 5 \ln{\left(x \right)} - 5\right)\right)|_{\left(x=4\right)}=-40 - 16 \ln{\left(4 \right)}^{2} + 40 \ln{\left(4 \right)}$$$
$$$\left(2 x \left(- 2 \ln{\left(x \right)}^{2} + 5 \ln{\left(x \right)} - 5\right)\right)|_{\left(x=2\right)}=-20 - 8 \ln{\left(2 \right)}^{2} + 20 \ln{\left(2 \right)}$$$
$$$\int_{2}^{4}\left( - 4 \ln{\left(x \right)}^{2} + 2 \ln{\left(x \right)} \right)dx=\left(2 x \left(- 2 \ln{\left(x \right)}^{2} + 5 \ln{\left(x \right)} - 5\right)\right)|_{\left(x=4\right)}-\left(2 x \left(- 2 \ln{\left(x \right)}^{2} + 5 \ln{\left(x \right)} - 5\right)\right)|_{\left(x=2\right)}=- 16 \ln{\left(4 \right)}^{2} - 20 - 20 \ln{\left(2 \right)} + 8 \ln{\left(2 \right)}^{2} + 40 \ln{\left(4 \right)}$$$
Answer: $$$\int_{2}^{4}\left( - \ln{\left(x^{2} \right)}^{2} + \ln{\left(x^{2} \right)} \right)dx=- 16 \ln{\left(4 \right)}^{2} - 20 - 20 \ln{\left(2 \right)} + 8 \ln{\left(2 \right)}^{2} + 40 \ln{\left(4 \right)}\approx -5.31653794582256$$$