Can you solve it?

Putting $\ln x=t$

$\large \displaystyle \int\Big( \ln t+\dfrac{1}{t^2}\Big)e^tdt$

$\large \displaystyle \int\Bigg(\Big( \ln t+\dfrac{1}{t}\Big)e^tdt-\Big(\dfrac{1}{t}-\dfrac{1}{t^2}\Big)e^tdt\Bigg)$

$\large \displaystyle \Bigg(\Big(\ln(t) e^t\Big)-\Big(\dfrac{e^t}{t}\Big)\Bigg)$ $\large \displaystyle \Big( \text{using} \int(f(x)+f'(x))e^x=e^x(f(x))\Big)$

Replacing $t$ with $\ln x$ ,

$\large \displaystyle \color{#3D99F6}{\boxed{f'(x)=x \ln (\ln x)-\dfrac{x}{\ln x}}}$