To evaluate the integral $\int_0^1 (x+4)(x-1)^4 \, dx$ using substitution, let’s proceed with the following steps:

1. Choose a Substitution: Let $u = x - 1$. Then $du = dx$, and we need to adjust the limits according to this substitution.

2. Change the Limits:

   • When $x = 0$: $u = 0 - 1 = -1$.
   • When $x = 1$: $u = 1 - 1 = 0$.

   So, the integral becomes: $\int_{-1}^0 (u + 5) u^4 \, du$ (since $x = u + 1$, $x + 4 = u + 5$).

3. Rewrite the Integral: Expanding $(u + 5) u^4$: $\int_{-1}^0 (u + 5) u^4 \, du = \int_{-1}^0 (u^5 + 5u^4) \, du$

4. Integrate Term by Term:

   • $\int u^5 \, du = \frac{u^6}{6}$
   • $\int 5u^4 \, du = 5 \cdot \frac{u^5}{5} = u^5$

   So the integral becomes: $\left[ \frac{u^6}{6} + u^5 \right]_{-1}^0$

5. Evaluate the Limits:

   • At $u = 0$: $\frac{0^6}{6} + 0^5 = 0$
   • At $u = -1$: $\frac{(-1)^6}{6} + (-1)^5 = \frac{1}{6} - 1 = -\frac{5}{6}$

   Therefore, the integral is: $0 - \left(-\frac{5}{6}\right) = \frac{5}{6}$

The final answer is $\frac{5}{6}$.

Would you like further details on any of these steps or have any questions?

Here are five related questions for further practice:

1. How would you apply substitution to an integral with different bounds?
2. How do you choose an appropriate substitution in integration?
3. Can you solve an integral without substitution if it's complex?
4. What are the general rules for changing bounds in definite integrals?
5. How would you integrate $\int (x+4)(x-2)^4 \, dx$ using substitution?

Tip: Always check if simplifying the integrand or using substitution makes the integration process easier.