Can someone help me with this?
$\displaystyle \begin{align*}\log(e^{2t}+5) & = \log \left[ 5 \left( \dfrac{ e^{2t} }{5} + 1 \right) \right] \\ & = \log(5) + \log\left( \dfrac{ e^{2t} }{5} + 1 \right)\end{align*}$
So, $\displaystyle \int_2^x \log(e^{2t}+5)dt = (x-2)\log(5) + \int_2^x \left( \sum_{n\ge 1} (-1)^{n+1}\dfrac{e^{2nt}}{n5^n} \right)dt$
Note: I replaced $\displaystyle \log\left(\dfrac{e^{2t}}{5}+1\right)$ with its power series. The radius of convergence of the power series is $\displaystyle \left| \dfrac{e^{2t} }{5} \right| < 1$, but over the complex numbers, it can be analytically extended uniquely to the original function. Similarly, once you integrate, the radius of convergence of the series you obtain will also only be finite, but can also be analytically extended (uniquely) over the complex numbers to an expression containing the dilogarithm function (the polylogarithm function where $\displaystyle n=2$).