Answer:
13
Solution: Using the fact that $x=e^{\ln (x)}$, we evaluate the integral as follows:
$$
\begin{aligned}
\int x^{2 x^{2}+1}+\ln \left(x^{2 x^{2 x^{2}+1}}\right) d x & =\int x^{2 x^{2}+1}+x^{2 x^{2}+1} \ln \left(x^{2}\right) d x \\
& =\int e^{\ln (x)\left(2 x^{2}+1\right)}\left(1+\ln \left(x^{2}\right)\right) d x \\
& =\int x e^{x^{2} \ln \left(x^{2}\right)}\left(1+\ln \left(x^{2}\right)\right) d x
\end{aligned}
$$

Noticing that the derivative of $x^{2} \ln \left(x^{2}\right)$ is $2 x\left(1+\ln \left(x^{2}\right)\right)$, it follows that the integral evaluates to
$$
\frac{1}{2} e^{x^{2} \ln \left(x^{2}\right)}=\frac{1}{2} x^{2 x^{2}} .
$$

Evaluating this from 1 to $\sqrt{3}$ we obtain the answer.
8. $[7]$ Compute
$$
\int_{1}^{\sqrt{3}} x^{2 x^{2}+1}+\ln \left(x^{2 x^{2 x^{2}+1}}\right) d x \text {. }
$$