I want to prove the divergence of: $$\int_{0}^{\infty} \frac{1}{x^2-6x+8}\mathrm dx$$
$$\int_{0}^{\infty} \frac{1}{x^2-6x+8}\mathrm dx=\int_{4}^{\infty}\frac{1}{x^2-6x+8}\mathrm dx +\int_{2}^{4} \frac{1}{x^2-6x+8}\mathrm dx+\int_{0}^{2} \frac{1}{x^2-6x+8}\mathrm dx$$
- $\int_{4}^{\infty} \frac{1}{x^2-6x+8}\mathrm dx$ diverges to $+\infty$
- $\int_{2}^{4} \frac{1}{x^2-6x+8}\mathrm dx$ diverges to $-\infty$
- $\int_{0}^{2} \frac{1}{x^2-6x+8}\mathrm dx$ diverges to $+\infty$
How can I prove that the sum of theese three integrals also diverges?
You can first write
$\int_{0}^{\infty} \frac{1}{x^2-6x+8}\mathrm dx=\int_{0}^{2}\frac{1}{x^2-6x+8}\mathrm dx +\int_{2}^{3} \frac{1}{x^2-6x+8}\mathrm dx+\int_{3}^{4} \frac{1}{x^2-6x+8}\mathrm dx+\int_{4}^{\infty}\frac{1}{x^2-6x+8}\mathrm dx$.
Then you can show that any one of the first 3 integrals diverges, using that $\int\frac{1}{x^2-6x+8}\;dx=\frac{1}{2}\ln\lvert\frac{x-4}{x-2}\rvert+C$, to conclude that $\int_0^{\infty}\frac{1}{x^2-6x+8}\;dx$ diverges.
For example, $\int_0^2\frac{1}{x^2-6x+8}\;dx=\lim_{t\to2^{-}}\int_0^{t}\frac{1}{x^2-6x+8}\;dx=\lim_{t\to2^{-}}\frac{1}{2}\left[\ln\lvert\frac{t-4}{t-2}\rvert-\ln2\right]=\infty$,
so $\int_0^2\frac{1}{x^2-6x+8}\;dx$ diverges.