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

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Calculate the integral $$ \int \frac{3x^3+x}{1+x^4} \, dx $$.
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int \frac{3x^{3}+x}{1+x^{4}} dx$$
- step1: Rewrite the fraction:
  $$\int \left(\frac{3x^{3}}{1+x^{4}}+\frac{x}{1+x^{4}}\right) dx$$
- step2: Use properties of integrals:
  $$\int \frac{3x^{3}}{1+x^{4}} dx+\int \frac{x}{1+x^{4}} dx$$
- step3: Evaluate the integral:
  $$\frac{3}{4}\ln{\left(x^{4}+1\right)}+\int \frac{x}{1+x^{4}} dx$$
- step4: Evaluate the integral:
  $$\frac{3}{4}\ln{\left(x^{4}+1\right)}+\frac{1}{2}\arctan\left(x^{2}\right)$$
- step5: Add the constant of integral C:
  $$\frac{3}{4}\ln{\left(x^{4}+1\right)}+\frac{1}{2}\arctan\left(x^{2}\right) + C, C \in \mathbb{R}$$
The integral of $$ \frac{3x^{3}+x}{1+x^{4}} $$ with respect to $$ x $$ is $$ \frac{3}{4}\ln{(x^{4}+1)}+\frac{1}{2}\arctan(x^{2}) + C $$, where $$ C $$ is an arbitrary constant.
The integral of $$ \frac{3x^{3}+x}{1+x^{4}} $$ with respect to $$ x $$ is $$ \frac{3}{4}\ln{(x^{4}+1)}+\frac{1}{2}\arctan(x^{2}) + C $$.

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

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