The value of $\int\left(3 x^2 \tan \frac{1}{x}-x \sec ^2 \frac{1}{x}\right) d x$, is

$x^3 \tan \frac{1}{x}+C$

Let

$I=\int\left(3 x^2 \tan \frac{1}{x}-x \sec ^2 \frac{1}{x}\right) d x$

$\Rightarrow I =\int 3 x^2 \tan \frac{1}{x} d x-\int x \sec ^2 \frac{1}{x} d x$

$\Rightarrow I =x^3 \tan \frac{1}{x}+\int x \sec ^2 \frac{1}{x} d x-\int x \sec ^2 \frac{1}{x} d x+C$

$=x^3 \tan \frac{1}{x}+C$