Evaluate the following integrals:

Assume $x+\log x \sec x=t$

$d(x+\log x \sec x)=d t$

$1+\frac{\sec x \tan x}{\sec x} d x=d t$

$(1+\tan x) d x=d t$

Put $\mathrm{t}$ and $\mathrm{dt}$ in given equation we get

$\Rightarrow \int \frac{\mathrm{d} t}{t}$

$=\ln |t|+c$

But $t=x+\log x \sec x$

$=\ln |x+\log x \sec x|+c$