$\int\{\sin (\log x)+\cos (\log x)\} d x$

$x \sin (\log x)+c$ $x \cos (\log x)+c$ $x \log (\sin x)+c$ $x \log (\cos x)+c$

$x \sin (\log x)+c$

Put log x = t ⇒ dx = et dt x = et $I=\int e^t(\sin t+\cos t) d t=e^{t} \sin t=x \sin (\log x)+c$ Hence (1) is the correct answer.