To solve the integral \(\int t^{13} e^{-t^7} \, dt\), we can use a substitution method.

### Step 1: Substitution
Let's substitute \( u = t^7 \). Then, the derivative of \( u \) with respect to \( t \) is:
\[
\frac{du}{dt} = 7t^6
\]
This implies:
\[
du = 7t^6 \, dt
\]

Now, rewrite the integral by isolating \( t^6 \, dt \) in the substitution:
\[
\int t^{13} e^{-t^7} \, dt = \int t^{13} e^{-u} \, \frac{du}{7t^6}
\]

### Step 2: Simplify the Integral
Notice that \( t^{13} = t^7 \cdot t^6 = u \cdot t^6 \). Therefore, the integral becomes:
\[
\int t^{13} e^{-t^7} \, dt = \frac{1}{7} \int u \cdot e^{-u} \, du
\]

### Step 3: Integration by Parts
To solve \(\int u \cdot e^{-u} \, du\), we use integration by parts. Let:
- \( v = u \)
- \( dw = e^{-u} \, du \)

Then:
- \( dv = du \)
- \( w = -e^{-u} \)

Now apply the integration by parts formula:
\[
\int u \cdot e^{-u} \, du = -u \cdot e^{-u} + \int e^{-u} \, du
\]
\[
\int u \cdot e^{-u} \, du = -u \cdot e^{-u} - e^{-u} + C
\]
\[
\int u \cdot e^{-u} \, du = -(u + 1) \cdot e^{-u} + C
\]

### Step 4: Back Substitute
Now, substitute back \( u = t^7 \):
\[
\int t^{13} e^{-t^7} \, dt = \frac{-1}{7} \cdot (t^7 + 1) \cdot e^{-t^7} + C
\]

### Final Answer
\[
\int t^{13} e^{-t^7} \, dt = -\frac{1}{7} (t^7 + 1) e^{-t^7} + C
\]

Would you like further details on any of the steps, or do you have any questions?

Here are 5 related questions you might find interesting:

1. How does the method of substitution work in different types of integrals?
2. Can we always use integration by parts for integrals involving a product of functions?
3. How do we choose the appropriate substitution for an integral?
4. What are some other examples of integrals that involve exponential functions?
5. How does the constant of integration, \(C\), affect the final answer in indefinite integrals?

**Tip:** Always check if a simpler substitution can simplify an integral before moving to more complex methods like integration by parts.

Learn how to solve the integral ∫ t^13 e^(-t^7) dt using the substitution method and integration by parts. This problem involves advanced calculus concepts and is suitable for undergraduate students studying calculus. Get a detailed solution and explanations.

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