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

$x^3 \tan \frac{1}{x}+C$ $x^2 \tan \frac{1}{x}+C$ $x \tan \frac{1}{x}+c$ none of these

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

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