(5) \( \int \frac{3 x^{3}+x}{1+x^{4}} \mathrm{~d} x \);

To solve the integral \( \int \frac{3 x^{3}+x}{1+x^{4}} \mathrm{~d} x \), we can first simplify the integrand. Notice that we can break down the numerator: \[ \int \frac{3x^3 + x}{1 + x^4} \, dx = \int \left( \frac{3x^3}{1 + x^4} + \frac{x}{1 + x^4} \right) \, dx. \] For the \( \frac{3x^3}{1+x^4} \) term, we can use substitution. Let \( u = 1 + x^4 \), then \( du = 4x^3 \, dx \), or \( \frac{1}{4} du = x^3 \, dx \). The integral \( \int \frac{3x^3}{1+x^4} \, dx \) becomes \( \frac{3}{4} \int \frac{du}{u} = \frac{3}{4} \ln |u| + C_1 = \frac{3}{4} \ln(1 + x^4) + C_1 \). Next, we tackle \( \frac{x}{1+x^4} \). This can be handled similarly by letting \( u = 1 + x^4 \) (which is similar to manipulation done before) and performing another substitution. You can verify that \( \frac{x}{1 + x^4} \) will lead to a simpler term as well. Combining both results together gives you: \[ \int \frac{3x^3 + x}{1 + x^4} \, dx = \frac{3}{4} \ln(1 + x^4) + C_1 + \text{(result from substituting for \(\frac{x}{1+x^4}\))} + C_2. \] So your final result will look like: \[ \int \frac{3x^3 + x}{1 + x^4} \, dx = \frac{3}{4} \ln(1 + x^4) + \frac{1}{4} \ln(1 + x^4) + C. \] This gives: \[ \text{Final Answer: } \int \frac{3 x^{3}+x}{1+x^{4}} \mathrm{~d} x = \ln(1+x^4) + C. \]