$$\int f(x) d x .$$

【解】由 $\int x^3 f^{\prime}(x) \mathrm{d} x=x^2 \cos x-4 x \sin x-6 \cos x+C$ 两边对 $x$ 求导得, $x^3 f^{\prime}(x)=2 \sin x-2 x \cos x-x^2 \sin x$, 即 $f^{\prime}(x)=\frac{2}{x^3} \sin x-\frac{2}{x^2} \cos x-\frac{1}{x} \sin x$.

\begin{aligned} f(x) & =\int \frac{2}{x^3} \sin x \mathrm{~d} x-\int \frac{2}{x^2} \cos x \mathrm{~d} x-\int \frac{1}{x} \sin x \mathrm{~d} x \\ & =-\int \sin x \mathrm{~d}\left(\frac{1}{x^2}\right)-\int \frac{2}{x^2} \cos x \mathrm{~d} x-\int \frac{1}{x} \sin x \mathrm{~d} x \\ & =-\frac{1}{x^2} \sin x-\int \frac{1}{x^2} \cos x \mathrm{~d} x-\int \frac{1}{x} \sin x \mathrm{~d} x \\ & =-\frac{1}{x^2} \sin x+\int \cos x \mathrm{~d}\left(\frac{1}{x}\right)-\int \frac{1}{x} \sin x \mathrm{~d} x \\ & =-\frac{1}{x^2} \sin x+\frac{1}{x} \cos x+\int \frac{1}{x} \sin x \mathrm{~d} x-\int \frac{1}{x} \sin x \mathrm{~d} x \\ & =-\frac{1}{x^2} \sin x+\frac{1}{x} \cos x+C . \end{aligned}

\begin{aligned} \int f(x) \mathrm{d} x & =\int\left(-\frac{1}{x^2} \sin x+\frac{1}{x} \cos x\right) \mathrm{d} x=\int \sin x \mathrm{~d}\left(\frac{1}{x}\right)+\int \frac{1}{x} \cos x \mathrm{~d} x \\ & =\frac{1}{x} \sin x-\int \frac{1}{x} \cos x \mathrm{~d} x+\int \frac{1}{x} \cos x \mathrm{~d} x=\frac{1}{x} \sin x+C_1, \end{aligned}

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