Use substitution to find the indefinite integral.
\[
\int \frac{x^{3}+2 x}{x^{4}+4 x^{2}-1} d x
\]

Step 1 ：Given the integral \(\int \frac{x^{3}+2 x}{x^{4}+4 x^{2}-1} d x\)

Step 2 ：Let's use substitution to simplify the integral. Let \(u = x^{4} + 4x^{2} - 1\), then \(du = (4x^{3} + 8x)dx\)

Step 3 ：However, the numerator is not exactly the derivative of the denominator. We can rewrite the numerator as \(x^{3} + 2x = \frac{1}{4}(4x^{3} + 8x) - x\)

Step 4 ：This allows us to express the integral in terms of \(u\) and \(du\)

Step 5 ：Substituting back into the original integral, we get \(\int \frac{1}{4} du - \int x dx\)

Step 6 ：Integrating, we get \(\frac{1}{4}u - 0.176776695296637\arctan(0.707106781186547x)\)

Step 7 ：Substituting \(u\) back in terms of \(x\), we get \(\frac{1}{4}x - 0.176776695296637\arctan(0.707106781186547x)\)

Step 8 ：Adding the constant of integration \(C\), the final answer is \(\boxed{\frac{1}{4}x - 0.176776695296637\arctan(0.707106781186547x) + C}\)