$I=\int \frac{-3 x}{\sqrt{8-x^{2}}} d x$

Step 1 ：Let \(u = 8 - x^2\)

Step 2 ：Differentiate \(u\) with respect to \(x\) to find \(du/dx\), \(du/dx = -2x\)

Step 3 ：Rearrange to find \(dx\) in terms of \(du\), \(dx = -du/(2x)\)

Step 4 ：Substitute \(u\) and \(dx\) into the integral, \(I = \int \frac{-3x}{\sqrt{u}} \cdot -du/(2x)\)

Step 5 ：Simplify the integral, \(I = \frac{3}{2} \int u^{-1/2} du\)

Step 6 ：Use the power rule for integration, \(I = \frac{3}{2} \cdot 2u^{1/2} + C\)

Step 7 ：Substitute \(u\) back into the integral, \(I = 3\sqrt{8 - x^2} + C\)

Step 8 ：Check the result. Differentiating \(3\sqrt{8 - x^2} + C\) with respect to \(x\) gives \(-3x/\sqrt{8 - x^2}\), confirming the solution is correct

Step 9 ：\(\boxed{I = 3\sqrt{8 - x^2} + C}\)