The value of $\int\left(3 x^2 \tan \frac{1}{x}-x \sec ^2 \frac{1}{x}\right) d x$, is
Answer & explanation
Correct answer: option 1
Let
$I=\int\left(3 x^2 \tan \frac{1}{x}-x \sec ^2 \frac{1}{x}\right) d x$
$\Rightarrow I =\int 3 x^2 \tan \frac{1}{x} d x-\int x \sec ^2 \frac{1}{x} d x$
$\Rightarrow I =x^3 \tan \frac{1}{x}+\int x \sec ^2 \frac{1}{x} d x-\int x \sec ^2 \frac{1}{x} d x+C$
$=x^3 \tan \frac{1}{x}+C$