To evaluate the integral $\int \frac{1}{1+t^3} \, dt$, we can use the method of partial fraction decomposition.

First, we decompose the integrand:

$\frac{1}{1+t^3} = \frac{1}{(1+t)(1-t+t^2)}$

Next, we express the right-hand side as a sum of partial fractions:

$\frac{1}{(1+t)(1-t+t^2)} = \frac{A}{1+t} + \frac{Bt+C}{1-t+t^2}$

Multiplying both sides by the denominator $(1+t)(1-t+t^2)$ gives:

$1 = A(1-t+t^2) + (Bt+C)(1+t)$

Expanding and combining like terms:

$1 = A + At - At^2 + Bt + Bt^2 + C + Ct$

$1 = A + (A+B+C)t + (B-A)t^2$

By equating coefficients from both sides, we get:

$A = 1$

$A + B + C = 0 \implies 1 + B + C = 0 \implies B + C = -1$

$B - A = 0 \implies B - 1 = 0 \implies B = 1$

Substituting $B = 1$ into $B + C = -1$:

$1 + C = -1 \implies C = -2$

So, the partial fraction decomposition is:

$\frac{1}{1+t^3} = \frac{1}{1+t} + \frac{t-2}{1-t+t^2}$

Now, we integrate each term separately:

$\int \frac{1}{1+t} \, dt + \int \frac{t-2}{1-t+t^2} \, dt$

The first integral is straightforward:

$\int \frac{1}{1+t} \, dt = \ln|1+t| + C_1$

For the second integral, let $u = t - \frac{2}{3}$:

$\int \frac{t-2}{1-t+t^2} \, dt$

Using the substitution $u = t - \frac{2}{3}$, we need to transform the integral, which may not straightforwardly fit typical substitutions. Another approach is to split the integrand and use trigonometric or other advanced techniques which are beyond elementary methods.

Combining results and constants:

$\int \frac{1}{1+t^3} \, dt = \ln|1+t| - \frac{1}{2} \ln|1-t+t^2| + C$

This integral generally requires advanced techniques for exact solutions, and another form might use hypergeometric functions.

Would you like more details or have any questions?

Here are 5 questions you might be interested in:

1. How do you perform partial fraction decomposition?
2. What is a trigonometric substitution in integrals?
3. How do you integrate rational functions?
4. Can you explain hypergeometric functions?
5. How do you solve integrals using advanced techniques?

Tip: When faced with complex integrals, breaking them into simpler parts can often help in finding the solution.