Given that $f(x) = \frac{x^2}{x-2}$ show that $8 \le \int_3^4 f(x)\,dx \le 9$. Consider the function:
$$f : \mathbb{R} \setminus \{ 2\} \rightarrow \mathbb{R} \hspace{2cm} f(x) = \dfrac{x^2}{x - 2}$$
I have to show that the following statement is true:
$$8 \le \int_3^4 f(x) \, dx \le 9$$
The first thing I did was to find:
$$\int_3^4 f(x) \, dx = \frac{11}{2} + 4 \ln 2$$
I think this is correct. The problem I have now is to show the following:
$$8 \le \dfrac{11}{2} + 4 \ln 2 \le 9$$
What I did in order to accomplish this was to first show that the result of the integral is first $\ge 8$ and then that it is $\le 9$. So first I tried to show:
$$\frac{11}{2} + 4 \ln 2 \ge 8$$
$$\frac{11}{2} - 8 + 4 \ln 2 \ge 0$$
After a few calculations I got to the point where I have to show:
$$8 \ln 2 \ge 5$$
$$\ln 2^8 \ge \ln e^5$$
$$ 256 \ge e^5 \tag 1$$
So I have to show that $256 \ge e ^5$. In a similar manner, working with the other part of the inequality (to trying to show that the result of the integral is $\le 9$) I have to show:
$$ 256 \le e^7 \tag 2$$
So I am stuck at proving $(1)$ and $(2)$. If I approximate $e \approx 2.71$ there is no problem, the inequalities are clearly true, but this feels a bit sloppy. Is there another way to prove $(1)$ and $(2)$ (or even the whole given integral inequality from the start) a little more rigorously, not relying on approximations?

A very quick way to show this is by finding the area under the curve in the interval from 3 to 4
The plot of $ f(x) $ looks like this- here is the plot.
From the plot it's very clear that area under the curve in $ [3, 4] $ is bounded between 8 and 9
Also, this function can be re-written as:
$$
f(x)=\frac{x^{2}}{x-2} = \frac{x^{2}-4+4}{x-2}= \frac{x^{2}-4}{x-2} + \frac{4}{x-2} \
=x+2+ \frac{4}{x-2}$$
For the values of $ f(x) $ in $ [3,4], f(3)=5+4=9 $ and $ f(4)= 6+2=8$,  function integral in the range $[3,4]$ is calculated as
$$\begin{aligned}
f(x)&=\frac{x^{2}}{x-2}\\
&= \frac{x^{2}-4+4}{x-2}\\
&= \frac{x^{2}-4}{x-2} + \frac{4}{x-2}\\
&=x+2+ \frac{4}{x-2}\end{aligned}$$
$$\begin{aligned}
&\int_{3}^{4} x+2+ \frac{4}{x-2}\\
=& \left[\frac{x^{2}}{2}+2x+4\ln(x-2)\right]_3^4\\
=& \left[\frac{4^{2}}{2}+8+4\ln2\right]-\left[\frac{3^{2}}{2}+6+0\right] \\=& 16+ 4\ln2-4.5-6 \\
=& 5.5+4\ln2 (\because \ln2\approx .69)\\ 
\approx& 8.3
\end{aligned}$$
