$\int\left[\frac{(1+\log x)}{\cos ^{2}(x \log x)}\right] d x=$
$\int\left[\frac{(1+\log x)}{\cos ^{2}(x \log x)}\right] d x=$
$\sin (x \log x)+c$
$\sin ^{2}(x \log x)+c$
$\log (x \log x)+c$
$\tan (x \log x)+c$
Solution
$I=\int \frac{(1+\log x)}{\cos ^{2}(x \log x)} d x$
Put $x \log x=t \Rightarrow\left[x \cdot \frac{1}{x}+\log x(1)\right] d x=d t \Rightarrow(1-\log x) d x=d t$
$1=\int \frac{1}{\cos ^{2} t} d t=\int \sec ^{2} t d t=\tan t+c$
$=\tan (x \log x)-c$