Solution
= Solution
Let $H(u)=\sin^2u+\ln u$. The <chain rule> gives $H'(u)=\sin2u+1/u$, so the two given summands combine into its <logarithmic derivative>, $f(u)=H'(u)/H(u)$. For $u>1$, $H(u)>0$, so the real logarithm is defined throughout the integration path. Since $H(\pi)=\ln\pi$,
$$
\boxed{F(x)=\ln\left(\frac{\sin^2x+\ln x}{\ln\pi}\right),\qquad x>1.}
$$
This satisfies $F(\pi)=0$, and <differentiation> recovers both terms of the integrand.