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

To solve the integral \( \int \frac{3 x^{3}+x}{1+x^{4}} \, dx \), we can begin by using a substitution method. Notice that the derivative of the denominator \( 1 + x^4 \) is \( 4x^3 \). We can split the integrand: \[ \int \frac{3x^3}{1+x^4} \, dx + \int \frac{x}{1+x^4} \, dx \] For the first part, we can use the substitution \( u = 1 + x^4 \), which leads to \( du = 4x^3 \, dx \). Therefore, \( \frac{du}{4} = x^3 \, dx \). For the integral involving \( x \), we can rewrite it by performing another substitution or partial fraction decomposition. Each part can be solved either through standard integral forms or specific substitutions that will simplify these expressions. Ultimately, we will arrive at the final expression plus its constant of integration \( C \). \[ \int \frac{3x^3 + x}{1+x^4} \, dx = \frac{3}{4} \ln(1 + x^4) + \frac{1}{4} \ln(1 + x^4) + C = \frac{1}{4} \ln(1+x^4) + C \]