5) \( \lim _{x \rightarrow+\infty}\left(x-\ln \left(x^{2}+e^{x}\right)\right)= \)

To evaluate the limit \( \lim _{x \rightarrow+\infty}\left(x-\ln \left(x^{2}+e^{x}\right)\right) \), let's analyze the expression inside the logarithm. As \( x \to +\infty \), \( e^x \) grows much faster than \( x^2 \), so we can approximate \( x^2 + e^x \) as \( e^x \). Thus, we have: \[ \ln(x^2 + e^x) \sim \ln(e^x) = x. \] Hence, we can rewrite the limit: \[ x - \ln(x^2 + e^x) \sim x - x = 0. \] Now, we need to determine the rate at which this limit approaches 0. We can express it as follows: \[ x - \ln(e^x (1 + \frac{x^2}{e^x})) = x - (x + \ln(1 + \frac{x^2}{e^x})), \] leading to: \[ -\ln(1 + \frac{x^2}{e^x}). \] Now, as \( x \to +\infty \): \[ \frac{x^2}{e^x} \to 0 \implies \ln(1 + \frac{x^2}{e^x}) \to 0. \] Thus, we have: \[ \lim _{x \rightarrow+\infty}\left(-\ln(1 + \frac{x^2}{e^x})\right) = 0. \] Now we conclude that: \[ \lim _{x \rightarrow+\infty}\left(x-\ln \left(x^{2}+e^{x}\right)\right) = 0. \] Therefore, the final answer is: \[ \boxed{0}. \]