Limit of an integral with cosine and logarithm

Rewriting the integrand. Note that for $t>1$,
$$\frac{\cos(\pi t)}{\ln t}=t\cos(\pi t)\cdot\frac{1}{t\ln t}.$$

Mean value theorem for integrals. On $[x,x^2]$ (with $1<x$), the function $t\mapsto t\cos(\pi t)$ is continuous and $t\mapsto\frac{1}{t\ln t}$ is continuous and positive. There exists $c_x\in[x,x^2]$ with
$$\int_x^{x^2}\frac{\cos(\pi t)}{\ln t}\,dt=c_x\cos(\pi c_x)\int_x^{x^2}\frac{dt}{t\ln t}.$$

Compute the remaining integral.
$$\int_x^{x^2}\frac{dt}{t\ln t}=\bigl[\ln(\ln t)\bigr]_x^{x^2}=\ln(\ln(x^2))-\ln(\ln x)=\ln 2.$$
So
$$\int_x^{x^2}\frac{\cos(\pi t)}{\ln t}\,dt=c_x\cos(\pi c_x)\ln 2.$$

Limit as $x\to1^+$. Since $c_x\in[x,x^2]$ and $x\to1^+$, we have $c_x\to1$. Thus $c_x\cos(\pi c_x)\to1\cdot\cos(\pi)=-1$. Therefore
$$\lim_{x\to1^+}\int_x^{x^2}\frac{\cos(\pi t)}{\ln t}\,dt=(-1)\cdot\ln 2=-\ln 2.$$