$\int 2 x\left(x^{2}+2\right)^{-4} d x=\square$

Step 1 ：Let \( u = x^2 + 2 \). Then, \( du = 2x dx \).

Step 2 ：Rewrite the integral in terms of \( u \): \( \int 2 x\left(x^{2}+2\right)^{-4} d x = \int u^{-4} du \)

Step 3 ：Integrate \( u^{-4} \) with respect to \( u \): \( \int u^{-4} du = \int u^{-4 + 1} \frac{1}{-4 + 1} du = \frac{1}{-3} u^{-3} \)

Step 4 ：Substitute back \( u = x^2 + 2 \): \( \frac{1}{-3} u^{-3} = \frac{1}{-3} (x^2 + 2)^{-3} \)

Step 5 ：\(\boxed{\int 2 x\left(x^{2}+2\right)^{-4} d x = \frac{1}{-3} (x^2 + 2)^{-3} + C}\)