To solve the integral

$\int x^3 (1 + x^4)^3 \, dx,$

we can use a substitution method. Let’s proceed step by step.

Step 1: Substitution

Let $u = 1 + x^4$. Then, differentiate both sides with respect to $x$:

$du = 4x^3 \, dx \quad \text{or} \quad \frac{du}{4} = x^3 \, dx.$

Step 2: Rewrite the integral

Substitute $u = 1 + x^4$ and $x^3 \, dx = \frac{du}{4}$ into the original integral:

$\int x^3 (1 + x^4)^3 \, dx = \int (u^3) \cdot \frac{du}{4}.$

This simplifies to:

$\frac{1}{4} \int u^3 \, du.$

Step 3: Integrate

Now integrate $u^3$:

$\frac{1}{4} \int u^3 \, du = \frac{1}{4} \cdot \frac{u^4}{4} = \frac{u^4}{16}.$

Step 4: Substitute back $u = 1 + x^4$

Finally, replace $u$ with $1 + x^4$:

$\frac{(1 + x^4)^4}{16} + C.$

Final Answer:

$\int x^3 (1 + x^4)^3 \, dx = \frac{(1 + x^4)^4}{16} + C.$

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

1. How would you solve $\int x^5 (1 + x^6)^2 \, dx$?
2. Can you solve $\int x^2 (1 + x^3)^5 \, dx$ using the same substitution method?
3. What happens if you use substitution in $\int (1 + x^2)^n \, dx$?
4. How do you differentiate $(1 + x^4)^4$ and verify the answer above?
5. Can you apply integration by parts to this problem instead?

Tip: When dealing with powers of binomials inside an integral, substitution is often the most efficient approach.