Calater to 3 d.p \[ \int \frac{x}{2 x^{2}-3} d x \]

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

Step 2 ：Let \(u = 2x^2 - 3\), then \(du = 4x dx\)

Step 3 ：Rewrite the integral in terms of u and du, we get \(\int \frac{1}{4} du\)

Step 4 ：Solve the integral, we get \(\frac{1}{4} \int du\)

Step 5 ：Simplify the integral, we get \(\frac{1}{4} u + C\)

Step 6 ：Substitute back for u, we get \(\frac{1}{4} (2x^2 - 3) + C\)

Step 7 ：\(\boxed{0.25 \log(2x^2 - 3) + C}\) is the final answer