What would be the answer to $\int (\cos(\frac1 x)+ \frac1 x\sin(\frac1 x))dx$ for $x \neq 0$?

Question

Well, that pretty much wraps up my question. But to give you some clues, I tried Geogebra and it said this Integral is undefined (I'm assuming since $x$ cannot be zero?). However, Symbolab does give me an answer $(x\cos(\frac1 x)+2C)$. However, the app fails to show the steps. Does anyone know if this answer is correct and, if so, how you go about solving the said integral? At first, I tried substituting $\frac1 x$, but that didn't seem to help much.
Thank you in advance for any help you can provide.

Answer 1

With $u=1/x$ it becomes arguably easier to spot:$$\int(\tfrac{-1}{u^2}\cos u+\tfrac{1}{u}\cdot-\sin u)du=\tfrac1u\cos u+C.$$Then there's integration by parts:$$\int\cos\tfrac1xdx=x\cos\tfrac1x-\int\tfrac1x\sin\tfrac1xdx.$$Either way though, you pretty much have to spot the same thing regardless of technique.

Answer 2

You can always check indefinite integrals solutions by differentiating them. So, Symbolab gives you right answer.

If you want to solve it by yourself, then first make substitution $u=1/x$, and then solve each of summands by partial integrations. Or, you can recognizine derivate of product under integral.