Answer
$$ C -(\frac{sin\frac{1}{x}}{x}+cos\frac{1}{x})$$
Work Step by Step
$$Let \,u=\frac{1}{x}\,,So\,du=-\frac{dx}{x^{2}}$$
$$\int \frac{cos\frac{1}{x}}{x^{3}}dx=-\int \frac{cosu}{\frac{1}{u}}du=-\int u \, cosu \,du$$
$$\int u \, cosu \,du=u\,sinu + C1 -\int sinu\,du$$
$$\int u \, cosu \,du=u\,sinu +cosu +C2$$
$$\int \frac{cos\frac{1}{x}}{x^{3}}dx=-\int u \, cosu \,du=C-u\,sinu -cosu$$
$$=C -(\frac{sin\frac{1}{x}}{x}+cos\frac{1}{x})$$
