show that the integral converges or that is diverges: $\int_{1}^{\infty}{\frac{2 + cos x}{x-1}dx} $ show that the integral converges or that is diverges: $$\int_{1}^{\infty}{\frac{2 + \cos x}{x-1}dx} $$ 
i know that it does converge from the solution, but i have no idea how to arrive at the conclusion. we are probably supposed to use p-integrals, but not sure how.
 A: Hint It diverges. How to see it? 
$$\frac{2+\cos x} {x-1} \geq \frac{1}{x-1} $$
A: For any $L>1$, both the integrals $\int_1^L \frac{2+\cos(x)}{x-1}\,dx$ and $\int_L^\infty \frac{2+\cos(x)}{x-1}\,dx$ diverge.
First, we write $\frac{2+\cos(x)}{x-1}=\frac{2-\cos(1)}{x-1}+\frac{\cos(x)-\cos(1)}{x-1}$. Therefore, we have for $\epsilon>0$
$$\begin{align}
\int_{1+\epsilon}^L \frac{2+\cos(x)}{x-1}\,dx&=\int_{1+\epsilon}^L \frac{2-\cos(1)}{x-1}\,dx+\int_{1+\epsilon}^L \frac{\cos(x)-\cos(1)}{x-1}\,dx\tag1
\end{align}$$
The first integral on the right-hand side of $(1)$ is equal to $(2-\cos(1))\left(\log(L-1)-\log(\epsilon)\right)$, which diverges as $\epsilon\to 0$, while the second integral has a removable discontinuity at $x=1$ and is therefore integrable.

For the integral $\int_L^\infty \frac{2+\cos(x)}{x-1}\,dx$ we see that $\int_L^\infty \frac{2}{x-1}\,dx$ diverges logarithmically, while the integral $\int_L^\infty \frac{\cos(x)}{x-1}\,dx$ converges (integrate by parts with $u=\frac1{x-1}$ and $v=\sin(x)$).  Hence, the sum of a convergent integral and a divergent integral is divergent.
